Solve x^2+(x+3/2)^2-8x-2(x+3/2)+7=0

$$x ^ { 2 } + ( \frac { x + 3 } { 2 } ) ^ { 2 } - 8 x - 2 ( \frac { x + 3 } { 2 } ) + 7 = 0$$

Solve for x

$$x=1$$

$$x=5$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2\left(x^{2}+\left(\frac{x+3}{2}\right)^{2}-8x-2\times \frac{x+3}{2}\right)+14=0$$

To raise \frac{x+3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$2\left(x^{2}+\frac{\left(x+3\right)^{2}}{2^{2}}-8x-2\times \frac{x+3}{2}\right)+14=0$$

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

$$2\left(\frac{\left(x^{2}-8x\right)\times 2^{2}}{2^{2}}+\frac{\left(x+3\right)^{2}}{2^{2}}-2\times \frac{x+3}{2}\right)+14=0$$

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

$$2\left(\frac{\left(x^{2}-8x\right)\times 2^{2}+\left(x+3\right)^{2}}{2^{2}}-2\times \frac{x+3}{2}\right)+14=0$$

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

$$2\left(\frac{4x^{2}-32x+x^{2}+6x+9}{2^{2}}-2\times \frac{x+3}{2}\right)+14=0$$

Combine like terms in 4x^{2}-32x+x^{2}+6x+9.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}-2\times \frac{x+3}{2}\right)+14=0$$

Express 2\times \frac{x+3}{2} as a single fraction.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}-\frac{2\left(x+3\right)}{2}\right)+14=0$$

Cancel out 2 and 2.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}-\left(x+3\right)\right)+14=0$$

To find the opposite of x+3, find the opposite of each term.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}-x-3\right)+14=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -x-3 times \frac{2^{2}}{2^{2}}.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}+\frac{\left(-x-3\right)\times 2^{2}}{2^{2}}\right)+14=0$$

Since \frac{5x^{2}-26x+9}{2^{2}} and \frac{\left(-x-3\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$2\times \frac{5x^{2}-26x+9+\left(-x-3\right)\times 2^{2}}{2^{2}}+14=0$$

Do the multiplications in 5x^{2}-26x+9+\left(-x-3\right)\times 2^{2}.

$$2\times \frac{5x^{2}-26x+9-4x-12}{2^{2}}+14=0$$

Combine like terms in 5x^{2}-26x+9-4x-12.

$$2\times \frac{5x^{2}-30x-3}{2^{2}}+14=0$$

Express 2\times \frac{5x^{2}-30x-3}{2^{2}} as a single fraction.

$$\frac{2\left(5x^{2}-30x-3\right)}{2^{2}}+14=0$$

Cancel out 2 in both numerator and denominator.

$$\frac{5x^{2}-30x-3}{2}+14=0$$

Divide each term of 5x^{2}-30x-3 by 2 to get \frac{5}{2}x^{2}-15x-\frac{3}{2}.

$$\frac{5}{2}x^{2}-15x-\frac{3}{2}+14=0$$

Add -\frac{3}{2} and 14 to get \frac{25}{2}.

$$\frac{5}{2}x^{2}-15x+\frac{25}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{2} for a, -15 for b, and \frac{25}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times \frac{5}{2}\times \frac{25}{2}}}{2\times \frac{5}{2}}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times \frac{5}{2}\times \frac{25}{2}}}{2\times \frac{5}{2}}$$

Multiply -4 times \frac{5}{2}.

$$x=\frac{-\left(-15\right)±\sqrt{225-10\times \frac{25}{2}}}{2\times \frac{5}{2}}$$

Multiply -10 times \frac{25}{2}.

$$x=\frac{-\left(-15\right)±\sqrt{225-125}}{2\times \frac{5}{2}}$$

Add 225 to -125.

$$x=\frac{-\left(-15\right)±\sqrt{100}}{2\times \frac{5}{2}}$$

Take the square root of 100.

$$x=\frac{-\left(-15\right)±10}{2\times \frac{5}{2}}$$

The opposite of -15 is 15.

$$x=\frac{15±10}{2\times \frac{5}{2}}$$

Multiply 2 times \frac{5}{2}.

$$x=\frac{15±10}{5}$$

Now solve the equation x=\frac{15±10}{5} when ± is plus. Add 15 to 10.

$$x=\frac{25}{5}$$

Divide 25 by 5.

$$x=5$$

Now solve the equation x=\frac{15±10}{5} when ± is minus. Subtract 10 from 15.

$$x=\frac{5}{5}$$

Divide 5 by 5.

$$x=1$$

The equation is now solved.

$$x=5 x=1$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$2\left(x^{2}+\left(\frac{x+3}{2}\right)^{2}-8x-2\times \frac{x+3}{2}\right)+14=0$$

To raise \frac{x+3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$2\left(x^{2}+\frac{\left(x+3\right)^{2}}{2^{2}}-8x-2\times \frac{x+3}{2}\right)+14=0$$

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

$$2\left(\frac{\left(x^{2}-8x\right)\times 2^{2}}{2^{2}}+\frac{\left(x+3\right)^{2}}{2^{2}}-2\times \frac{x+3}{2}\right)+14=0$$

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

$$2\left(\frac{\left(x^{2}-8x\right)\times 2^{2}+\left(x+3\right)^{2}}{2^{2}}-2\times \frac{x+3}{2}\right)+14=0$$

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

$$2\left(\frac{4x^{2}-32x+x^{2}+6x+9}{2^{2}}-2\times \frac{x+3}{2}\right)+14=0$$

Combine like terms in 4x^{2}-32x+x^{2}+6x+9.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}-2\times \frac{x+3}{2}\right)+14=0$$

Express 2\times \frac{x+3}{2} as a single fraction.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}-\frac{2\left(x+3\right)}{2}\right)+14=0$$

Cancel out 2 and 2.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}-\left(x+3\right)\right)+14=0$$

To find the opposite of x+3, find the opposite of each term.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}-x-3\right)+14=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -x-3 times \frac{2^{2}}{2^{2}}.

$$2\left(\frac{5x^{2}-26x+9}{2^{2}}+\frac{\left(-x-3\right)\times 2^{2}}{2^{2}}\right)+14=0$$

Since \frac{5x^{2}-26x+9}{2^{2}} and \frac{\left(-x-3\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$2\times \frac{5x^{2}-26x+9+\left(-x-3\right)\times 2^{2}}{2^{2}}+14=0$$

Do the multiplications in 5x^{2}-26x+9+\left(-x-3\right)\times 2^{2}.

$$2\times \frac{5x^{2}-26x+9-4x-12}{2^{2}}+14=0$$

Combine like terms in 5x^{2}-26x+9-4x-12.

$$2\times \frac{5x^{2}-30x-3}{2^{2}}+14=0$$

Express 2\times \frac{5x^{2}-30x-3}{2^{2}} as a single fraction.

$$\frac{2\left(5x^{2}-30x-3\right)}{2^{2}}+14=0$$

Cancel out 2 in both numerator and denominator.

$$\frac{5x^{2}-30x-3}{2}+14=0$$

Divide each term of 5x^{2}-30x-3 by 2 to get \frac{5}{2}x^{2}-15x-\frac{3}{2}.

$$\frac{5}{2}x^{2}-15x-\frac{3}{2}+14=0$$

Add -\frac{3}{2} and 14 to get \frac{25}{2}.

$$\frac{5}{2}x^{2}-15x+\frac{25}{2}=0$$

Subtract \frac{25}{2} from both sides. Anything subtracted from zero gives its negation.

$$\frac{5}{2}x^{2}-15x=-\frac{25}{2}$$

Divide both sides of the equation by \frac{5}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{2}x^{2}-15x}{\frac{5}{2}}=-\frac{\frac{25}{2}}{\frac{5}{2}}$$

Dividing by \frac{5}{2} undoes the multiplication by \frac{5}{2}.

$$x^{2}+\left(-\frac{15}{\frac{5}{2}}\right)x=-\frac{\frac{25}{2}}{\frac{5}{2}}$$

Divide -15 by \frac{5}{2} by multiplying -15 by the reciprocal of \frac{5}{2}.

$$x^{2}-6x=-\frac{\frac{25}{2}}{\frac{5}{2}}$$

Divide -\frac{25}{2} by \frac{5}{2} by multiplying -\frac{25}{2} by the reciprocal of \frac{5}{2}.

$$x^{2}-6x=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$