Solve {left(3x-4/8right)}^2+1={left(x-3x-4/8right)}^2+1

$${ \left( \frac{ 3x-4 }{ 8 } \right) }^{ 2 } +1= { \left(x- \frac{ 3x-4 }{ 8 } \right) }^{ 2 } +1$$

Solve for x

$$x=-4$$

$$x=0$$

Steps Using Factoring

To raise \frac{3x-4}{8} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x-4\right)^{2}}{8^{2}}+1=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{8^{2}}{8^{2}}.

$$\frac{\left(3x-4\right)^{2}}{8^{2}}+\frac{8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Since \frac{\left(3x-4\right)^{2}}{8^{2}} and \frac{8^{2}}{8^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(3x-4\right)^{2}+8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Do the multiplications in \left(3x-4\right)^{2}+8^{2}.

$$\frac{9x^{2}-24x+16+8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Combine like terms in 9x^{2}-24x+16+8^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x-\frac{3x-4}{8}\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+2x\left(-\frac{3x-4}{8}\right)+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Cancel out 8, the greatest common factor in 2 and 8.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{3x-4}{-4}x+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Multiply both numerator and denominator by -1.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{-3x+4}{4}x+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Calculate -\frac{3x-4}{8} to the power of 2 and get \left(\frac{3x-4}{8}\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{-3x+4}{4}x+\left(\frac{3x-4}{8}\right)^{2}+1$$

Express \frac{-3x+4}{4}x as a single fraction.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{\left(-3x+4\right)x}{4}+\left(\frac{3x-4}{8}\right)^{2}+1$$

To raise \frac{3x-4}{8} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{\left(-3x+4\right)x}{4}+\frac{\left(3x-4\right)^{2}}{8^{2}}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+1 times \frac{8^{2}}{8^{2}}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{\left(x^{2}+1\right)\times 8^{2}}{8^{2}}+\frac{\left(-3x+4\right)x}{4}+\frac{\left(3x-4\right)^{2}}{8^{2}}$$

Since \frac{\left(x^{2}+1\right)\times 8^{2}}{8^{2}} and \frac{\left(3x-4\right)^{2}}{8^{2}} have the same denominator, add them by adding their numerators.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{\left(x^{2}+1\right)\times 8^{2}+\left(3x-4\right)^{2}}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

Do the multiplications in \left(x^{2}+1\right)\times 8^{2}+\left(3x-4\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{64x^{2}+64+9x^{2}-24x+16}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

Combine like terms in 64x^{2}+64+9x^{2}-24x+16.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 8^{2} and 4 is 64. Multiply \frac{\left(-3x+4\right)x}{4} times \frac{16}{16}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x}{64}+\frac{16\left(-3x+4\right)x}{64}$$

Since \frac{73x^{2}+80-24x}{64} and \frac{16\left(-3x+4\right)x}{64} have the same denominator, add them by adding their numerators.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x+16\left(-3x+4\right)x}{64}$$

Do the multiplications in 73x^{2}+80-24x+16\left(-3x+4\right)x.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x-48x^{2}+64x}{64}$$

Combine like terms in 73x^{2}+80-24x-48x^{2}+64x.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{25x^{2}+80+40x}{64}$$

Calculate 8 to the power of 2 and get 64.

$$\frac{9x^{2}-24x+80}{64}=\frac{25x^{2}+80+40x}{64}$$

Divide each term of 9x^{2}-24x+80 by 64 to get \frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{25x^{2}+80+40x}{64}$$

Divide each term of 25x^{2}+80+40x by 64 to get \frac{25}{64}x^{2}+\frac{5}{4}+\frac{5}{8}x.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{25}{64}x^{2}+\frac{5}{4}+\frac{5}{8}x$$

Subtract \frac{25}{64}x^{2} from both sides.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}-\frac{25}{64}x^{2}=\frac{5}{4}+\frac{5}{8}x$$

Combine \frac{9}{64}x^{2} and -\frac{25}{64}x^{2} to get -\frac{1}{4}x^{2}.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{5}{4}+\frac{5}{8}x$$

Subtract \frac{5}{4} from both sides.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x+\frac{5}{4}-\frac{5}{4}=\frac{5}{8}x$$

Subtract \frac{5}{4} from \frac{5}{4} to get 0.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x=\frac{5}{8}x$$

Subtract \frac{5}{8}x from both sides.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x-\frac{5}{8}x=0$$

Combine -\frac{3}{8}x and -\frac{5}{8}x to get -x.

$$-\frac{1}{4}x^{2}-x=0$$

Factor out x.

$$x\left(-\frac{1}{4}x-1\right)=0$$

To find equation solutions, solve x=0 and -\frac{x}{4}-1=0.

$$x=0 x=-4$$

Steps Using the Quadratic Formula

To raise \frac{3x-4}{8} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x-4\right)^{2}}{8^{2}}+1=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{8^{2}}{8^{2}}.

$$\frac{\left(3x-4\right)^{2}}{8^{2}}+\frac{8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Since \frac{\left(3x-4\right)^{2}}{8^{2}} and \frac{8^{2}}{8^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(3x-4\right)^{2}+8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Do the multiplications in \left(3x-4\right)^{2}+8^{2}.

$$\frac{9x^{2}-24x+16+8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Combine like terms in 9x^{2}-24x+16+8^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x-\frac{3x-4}{8}\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+2x\left(-\frac{3x-4}{8}\right)+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Cancel out 8, the greatest common factor in 2 and 8.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{3x-4}{-4}x+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Multiply both numerator and denominator by -1.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{-3x+4}{4}x+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Calculate -\frac{3x-4}{8} to the power of 2 and get \left(\frac{3x-4}{8}\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{-3x+4}{4}x+\left(\frac{3x-4}{8}\right)^{2}+1$$

Express \frac{-3x+4}{4}x as a single fraction.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{\left(-3x+4\right)x}{4}+\left(\frac{3x-4}{8}\right)^{2}+1$$

To raise \frac{3x-4}{8} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{\left(-3x+4\right)x}{4}+\frac{\left(3x-4\right)^{2}}{8^{2}}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+1 times \frac{8^{2}}{8^{2}}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{\left(x^{2}+1\right)\times 8^{2}}{8^{2}}+\frac{\left(-3x+4\right)x}{4}+\frac{\left(3x-4\right)^{2}}{8^{2}}$$

Since \frac{\left(x^{2}+1\right)\times 8^{2}}{8^{2}} and \frac{\left(3x-4\right)^{2}}{8^{2}} have the same denominator, add them by adding their numerators.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{\left(x^{2}+1\right)\times 8^{2}+\left(3x-4\right)^{2}}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

Do the multiplications in \left(x^{2}+1\right)\times 8^{2}+\left(3x-4\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{64x^{2}+64+9x^{2}-24x+16}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

Combine like terms in 64x^{2}+64+9x^{2}-24x+16.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 8^{2} and 4 is 64. Multiply \frac{\left(-3x+4\right)x}{4} times \frac{16}{16}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x}{64}+\frac{16\left(-3x+4\right)x}{64}$$

Since \frac{73x^{2}+80-24x}{64} and \frac{16\left(-3x+4\right)x}{64} have the same denominator, add them by adding their numerators.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x+16\left(-3x+4\right)x}{64}$$

Do the multiplications in 73x^{2}+80-24x+16\left(-3x+4\right)x.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x-48x^{2}+64x}{64}$$

Combine like terms in 73x^{2}+80-24x-48x^{2}+64x.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{25x^{2}+80+40x}{64}$$

Calculate 8 to the power of 2 and get 64.

$$\frac{9x^{2}-24x+80}{64}=\frac{25x^{2}+80+40x}{64}$$

Divide each term of 9x^{2}-24x+80 by 64 to get \frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{25x^{2}+80+40x}{64}$$

Divide each term of 25x^{2}+80+40x by 64 to get \frac{25}{64}x^{2}+\frac{5}{4}+\frac{5}{8}x.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{25}{64}x^{2}+\frac{5}{4}+\frac{5}{8}x$$

Subtract \frac{25}{64}x^{2} from both sides.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}-\frac{25}{64}x^{2}=\frac{5}{4}+\frac{5}{8}x$$

Combine \frac{9}{64}x^{2} and -\frac{25}{64}x^{2} to get -\frac{1}{4}x^{2}.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{5}{4}+\frac{5}{8}x$$

Subtract \frac{5}{4} from both sides.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x+\frac{5}{4}-\frac{5}{4}=\frac{5}{8}x$$

Subtract \frac{5}{4} from \frac{5}{4} to get 0.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x=\frac{5}{8}x$$

Subtract \frac{5}{8}x from both sides.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x-\frac{5}{8}x=0$$

Combine -\frac{3}{8}x and -\frac{5}{8}x to get -x.

$$-\frac{1}{4}x^{2}-x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{4} for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\left(-\frac{1}{4}\right)}$$

Take the square root of 1.

$$x=\frac{-\left(-1\right)±1}{2\left(-\frac{1}{4}\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±1}{2\left(-\frac{1}{4}\right)}$$

Multiply 2 times -\frac{1}{4}.

$$x=\frac{1±1}{-\frac{1}{2}}$$

Now solve the equation x=\frac{1±1}{-\frac{1}{2}} when ± is plus. Add 1 to 1.

$$x=\frac{2}{-\frac{1}{2}}$$

Divide 2 by -\frac{1}{2} by multiplying 2 by the reciprocal of -\frac{1}{2}.

$$x=-4$$

Now solve the equation x=\frac{1±1}{-\frac{1}{2}} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{-\frac{1}{2}}$$

Divide 0 by -\frac{1}{2} by multiplying 0 by the reciprocal of -\frac{1}{2}.

$$x=0$$

The equation is now solved.

$$x=-4 x=0$$

Steps for Completing the Square

To raise \frac{3x-4}{8} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x-4\right)^{2}}{8^{2}}+1=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{8^{2}}{8^{2}}.

$$\frac{\left(3x-4\right)^{2}}{8^{2}}+\frac{8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Since \frac{\left(3x-4\right)^{2}}{8^{2}} and \frac{8^{2}}{8^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(3x-4\right)^{2}+8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Do the multiplications in \left(3x-4\right)^{2}+8^{2}.

$$\frac{9x^{2}-24x+16+8^{2}}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Combine like terms in 9x^{2}-24x+16+8^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\left(x-\frac{3x-4}{8}\right)^{2}+1$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x-\frac{3x-4}{8}\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+2x\left(-\frac{3x-4}{8}\right)+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Cancel out 8, the greatest common factor in 2 and 8.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{3x-4}{-4}x+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Multiply both numerator and denominator by -1.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{-3x+4}{4}x+\left(-\frac{3x-4}{8}\right)^{2}+1$$

Calculate -\frac{3x-4}{8} to the power of 2 and get \left(\frac{3x-4}{8}\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{-3x+4}{4}x+\left(\frac{3x-4}{8}\right)^{2}+1$$

Express \frac{-3x+4}{4}x as a single fraction.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{\left(-3x+4\right)x}{4}+\left(\frac{3x-4}{8}\right)^{2}+1$$

To raise \frac{3x-4}{8} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{9x^{2}-24x+80}{8^{2}}=x^{2}+\frac{\left(-3x+4\right)x}{4}+\frac{\left(3x-4\right)^{2}}{8^{2}}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+1 times \frac{8^{2}}{8^{2}}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{\left(x^{2}+1\right)\times 8^{2}}{8^{2}}+\frac{\left(-3x+4\right)x}{4}+\frac{\left(3x-4\right)^{2}}{8^{2}}$$

Since \frac{\left(x^{2}+1\right)\times 8^{2}}{8^{2}} and \frac{\left(3x-4\right)^{2}}{8^{2}} have the same denominator, add them by adding their numerators.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{\left(x^{2}+1\right)\times 8^{2}+\left(3x-4\right)^{2}}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

Do the multiplications in \left(x^{2}+1\right)\times 8^{2}+\left(3x-4\right)^{2}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{64x^{2}+64+9x^{2}-24x+16}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

Combine like terms in 64x^{2}+64+9x^{2}-24x+16.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x}{8^{2}}+\frac{\left(-3x+4\right)x}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 8^{2} and 4 is 64. Multiply \frac{\left(-3x+4\right)x}{4} times \frac{16}{16}.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x}{64}+\frac{16\left(-3x+4\right)x}{64}$$

Since \frac{73x^{2}+80-24x}{64} and \frac{16\left(-3x+4\right)x}{64} have the same denominator, add them by adding their numerators.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x+16\left(-3x+4\right)x}{64}$$

Do the multiplications in 73x^{2}+80-24x+16\left(-3x+4\right)x.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{73x^{2}+80-24x-48x^{2}+64x}{64}$$

Combine like terms in 73x^{2}+80-24x-48x^{2}+64x.

$$\frac{9x^{2}-24x+80}{8^{2}}=\frac{25x^{2}+80+40x}{64}$$

Calculate 8 to the power of 2 and get 64.

$$\frac{9x^{2}-24x+80}{64}=\frac{25x^{2}+80+40x}{64}$$

Divide each term of 9x^{2}-24x+80 by 64 to get \frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{25x^{2}+80+40x}{64}$$

Divide each term of 25x^{2}+80+40x by 64 to get \frac{25}{64}x^{2}+\frac{5}{4}+\frac{5}{8}x.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{25}{64}x^{2}+\frac{5}{4}+\frac{5}{8}x$$

Subtract \frac{25}{64}x^{2} from both sides.

$$\frac{9}{64}x^{2}-\frac{3}{8}x+\frac{5}{4}-\frac{25}{64}x^{2}=\frac{5}{4}+\frac{5}{8}x$$

Combine \frac{9}{64}x^{2} and -\frac{25}{64}x^{2} to get -\frac{1}{4}x^{2}.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x+\frac{5}{4}=\frac{5}{4}+\frac{5}{8}x$$

Subtract \frac{5}{8}x from both sides.

$$-\frac{1}{4}x^{2}-\frac{3}{8}x+\frac{5}{4}-\frac{5}{8}x=\frac{5}{4}$$

Combine -\frac{3}{8}x and -\frac{5}{8}x to get -x.

$$-\frac{1}{4}x^{2}-x+\frac{5}{4}=\frac{5}{4}$$

Subtract \frac{5}{4} from both sides.

$$-\frac{1}{4}x^{2}-x=\frac{5}{4}-\frac{5}{4}$$

Subtract \frac{5}{4} from \frac{5}{4} to get 0.

$$-\frac{1}{4}x^{2}-x=0$$

Multiply both sides by -4.

$$\frac{-\frac{1}{4}x^{2}-x}{-\frac{1}{4}}=\frac{0}{-\frac{1}{4}}$$

Dividing by -\frac{1}{4} undoes the multiplication by -\frac{1}{4}.

$$x^{2}+\left(-\frac{1}{-\frac{1}{4}}\right)x=\frac{0}{-\frac{1}{4}}$$

Divide -1 by -\frac{1}{4} by multiplying -1 by the reciprocal of -\frac{1}{4}.

$$x^{2}+4x=\frac{0}{-\frac{1}{4}}$$

Divide 0 by -\frac{1}{4} by multiplying 0 by the reciprocal of -\frac{1}{4}.

$$x^{2}+4x=0$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=2^{2}$$

Square 2.

$$x^{2}+4x+4=4$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+2=2 x+2=-2$$

Subtract 2 from both sides of the equation.

$$x=0 x=-4$$