Solve 10x^2+160=(-4x-8)^2

$$10 x ^ { 2 } + 160 = ( - 4 x - 8 ) ^ { 2 }$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=-12$$

Steps Using Factoring By Grouping

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

$$10x^{2}+160=16x^{2}+64x+64$$

Subtract 16x^{2} from both sides.

$$10x^{2}+160-16x^{2}=64x+64$$

Combine 10x^{2} and -16x^{2} to get -6x^{2}.

$$-6x^{2}+160=64x+64$$

Subtract 64x from both sides.

$$-6x^{2}+160-64x=64$$

Subtract 64 from both sides.

$$-6x^{2}+160-64x-64=0$$

Subtract 64 from 160 to get 96.

$$-6x^{2}+96-64x=0$$

Divide both sides by 2.

$$-3x^{2}+48-32x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}-32x+48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+48. To find a and b, set up a system to be solved.

$$a+b=-32 ab=-3\times 48=-144$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -144.

$$1,-144 2,-72 3,-48 4,-36 6,-24 8,-18 9,-16 12,-12$$

Calculate the sum for each pair.

$$1-144=-143 2-72=-70 3-48=-45 4-36=-32 6-24=-18 8-18=-10 9-16=-7 12-12=0$$

The solution is the pair that gives sum -32.

$$a=4 b=-36$$

Rewrite -3x^{2}-32x+48 as \left(-3x^{2}+4x\right)+\left(-36x+48\right).

$$\left(-3x^{2}+4x\right)+\left(-36x+48\right)$$

Factor out -x in the first and -12 in the second group.

$$-x\left(3x-4\right)-12\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(-x-12\right)$$

To find equation solutions, solve 3x-4=0 and -x-12=0.

$$x=\frac{4}{3} x=-12$$

Steps Using the Quadratic Formula

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

$$10x^{2}+160=16x^{2}+64x+64$$

Subtract 16x^{2} from both sides.

$$10x^{2}+160-16x^{2}=64x+64$$

Combine 10x^{2} and -16x^{2} to get -6x^{2}.

$$-6x^{2}+160=64x+64$$

Subtract 64x from both sides.

$$-6x^{2}+160-64x=64$$

Subtract 64 from both sides.

$$-6x^{2}+160-64x-64=0$$

Subtract 64 from 160 to get 96.

$$-6x^{2}+96-64x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-6x^{2}-64x+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -64 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-64\right)±\sqrt{\left(-64\right)^{2}-4\left(-6\right)\times 96}}{2\left(-6\right)}$$

Square -64.

$$x=\frac{-\left(-64\right)±\sqrt{4096-4\left(-6\right)\times 96}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-64\right)±\sqrt{4096+24\times 96}}{2\left(-6\right)}$$

Multiply 24 times 96.

$$x=\frac{-\left(-64\right)±\sqrt{4096+2304}}{2\left(-6\right)}$$

Add 4096 to 2304.

$$x=\frac{-\left(-64\right)±\sqrt{6400}}{2\left(-6\right)}$$

Take the square root of 6400.

$$x=\frac{-\left(-64\right)±80}{2\left(-6\right)}$$

The opposite of -64 is 64.

$$x=\frac{64±80}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{64±80}{-12}$$

Now solve the equation x=\frac{64±80}{-12} when ± is plus. Add 64 to 80.

$$x=\frac{144}{-12}$$

Divide 144 by -12.

$$x=-12$$

Now solve the equation x=\frac{64±80}{-12} when ± is minus. Subtract 80 from 64.

$$x=-\frac{16}{-12}$$

Reduce the fraction \frac{-16}{-12} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{3}$$

The equation is now solved.

$$x=-12 x=\frac{4}{3}$$

Steps for Completing the Square

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

$$10x^{2}+160=16x^{2}+64x+64$$

Subtract 16x^{2} from both sides.

$$10x^{2}+160-16x^{2}=64x+64$$

Combine 10x^{2} and -16x^{2} to get -6x^{2}.

$$-6x^{2}+160=64x+64$$

Subtract 64x from both sides.

$$-6x^{2}+160-64x=64$$

Subtract 160 from both sides.

$$-6x^{2}-64x=64-160$$

Subtract 160 from 64 to get -96.

$$-6x^{2}-64x=-96$$

Divide both sides by -6.

$$\frac{-6x^{2}-64x}{-6}=-\frac{96}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\left(-\frac{64}{-6}\right)x=-\frac{96}{-6}$$

Reduce the fraction \frac{-64}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{32}{3}x=-\frac{96}{-6}$$

Divide -96 by -6.

$$x^{2}+\frac{32}{3}x=16$$

Divide \frac{32}{3}, the coefficient of the x term, by 2 to get \frac{16}{3}. Then add the square of \frac{16}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{32}{3}x+\left(\frac{16}{3}\right)^{2}=16+\left(\frac{16}{3}\right)^{2}$$

Square \frac{16}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{32}{3}x+\frac{256}{9}=16+\frac{256}{9}$$

Add 16 to \frac{256}{9}.

$$x^{2}+\frac{32}{3}x+\frac{256}{9}=\frac{400}{9}$$

Factor x^{2}+\frac{32}{3}x+\frac{256}{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}.

$$\left(x+\frac{16}{3}\right)^{2}=\frac{400}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{16}{3}\right)^{2}}=\sqrt{\frac{400}{9}}$$

Simplify.

$$x+\frac{16}{3}=\frac{20}{3} x+\frac{16}{3}=-\frac{20}{3}$$

Subtract \frac{16}{3} from both sides of the equation.

$$x=\frac{4}{3} x=-12$$