Solve (x+7)^2=2x^2+8x+54

$$( x + 7 ) ^ { 2 } = 2 x ^ { 2 } + 8 x + 54$$

Solve for x

$$x=1$$

$$x=5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$x^{2}+14x+49=2x^{2}+8x+54$$

Subtract 2x^{2} from both sides.

$$x^{2}+14x+49-2x^{2}=8x+54$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+14x+49=8x+54$$

Subtract 8x from both sides.

$$-x^{2}+14x+49-8x=54$$

Combine 14x and -8x to get 6x.

$$-x^{2}+6x+49=54$$

Subtract 54 from both sides.

$$-x^{2}+6x+49-54=0$$

Subtract 54 from 49 to get -5.

$$-x^{2}+6x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-5\right)=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=5 b=1$$

Rewrite -x^{2}+6x-5 as \left(-x^{2}+5x\right)+\left(x-5\right).

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

Factor out -x in -x^{2}+5x.

$$-x\left(x-5\right)+x-5$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x+1\right)$$

To find equation solutions, solve x-5=0 and -x+1=0.

$$x=5 x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$x^{2}+14x+49=2x^{2}+8x+54$$

Subtract 2x^{2} from both sides.

$$x^{2}+14x+49-2x^{2}=8x+54$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+14x+49=8x+54$$

Subtract 8x from both sides.

$$-x^{2}+14x+49-8x=54$$

Combine 14x and -8x to get 6x.

$$-x^{2}+6x+49=54$$

Subtract 54 from both sides.

$$-x^{2}+6x+49-54=0$$

Subtract 54 from 49 to get -5.

$$-x^{2}+6x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-5\right)}}{2\left(-1\right)}$$

Multiply 4 times -5.

$$x=\frac{-6±\sqrt{36-20}}{2\left(-1\right)}$$

Add 36 to -20.

$$x=\frac{-6±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{-6±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±4}{-2}$$

Now solve the equation x=\frac{-6±4}{-2} when ± is plus. Add -6 to 4.

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

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-6±4}{-2} when ± is minus. Subtract 4 from -6.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=1 x=5$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$x^{2}+14x+49=2x^{2}+8x+54$$

Subtract 2x^{2} from both sides.

$$x^{2}+14x+49-2x^{2}=8x+54$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}+14x+49=8x+54$$

Subtract 8x from both sides.

$$-x^{2}+14x+49-8x=54$$

Combine 14x and -8x to get 6x.

$$-x^{2}+6x+49=54$$

Subtract 49 from both sides.

$$-x^{2}+6x=54-49$$

Subtract 49 from 54 to get 5.

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

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{5}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{5}{-1}$$

Divide 5 by -1.

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

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.

$$x=5 x=1$$