Solve (4-x)^2-(3x+2)^2=4(2-5x)

$$( 4 - x ) ^ { 2 } - ( 3 x + 2 ) ^ { 2 } = 4 ( 2 - 5 x )$$

Solve for x

$$x=\frac{\sqrt{2}}{2}\approx 0.707106781$$

$$x=-\frac{\sqrt{2}}{2}\approx -0.707106781$$

Steps by Finding Square Root

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

$$16-8x+x^{2}-\left(3x+2\right)^{2}=4\left(2-5x\right)$$

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

$$16-8x+x^{2}-\left(9x^{2}+12x+4\right)=4\left(2-5x\right)$$

To find the opposite of 9x^{2}+12x+4, find the opposite of each term.

$$16-8x+x^{2}-9x^{2}-12x-4=4\left(2-5x\right)$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$16-8x-8x^{2}-12x-4=4\left(2-5x\right)$$

Combine -8x and -12x to get -20x.

$$16-20x-8x^{2}-4=4\left(2-5x\right)$$

Subtract 4 from 16 to get 12.

$$12-20x-8x^{2}=4\left(2-5x\right)$$

Use the distributive property to multiply 4 by 2-5x.

$$12-20x-8x^{2}=8-20x$$

Add 20x to both sides.

$$12-20x-8x^{2}+20x=8$$

Combine -20x and 20x to get 0.

$$12-8x^{2}=8$$

Subtract 12 from both sides.

$$-8x^{2}=8-12$$

Subtract 12 from 8 to get -4.

$$-8x^{2}=-4$$

Divide both sides by -8.

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

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out -4.

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

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

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

$$16-8x+x^{2}-\left(3x+2\right)^{2}=4\left(2-5x\right)$$

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

$$16-8x+x^{2}-\left(9x^{2}+12x+4\right)=4\left(2-5x\right)$$

To find the opposite of 9x^{2}+12x+4, find the opposite of each term.

$$16-8x+x^{2}-9x^{2}-12x-4=4\left(2-5x\right)$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$16-8x-8x^{2}-12x-4=4\left(2-5x\right)$$

Combine -8x and -12x to get -20x.

$$16-20x-8x^{2}-4=4\left(2-5x\right)$$

Subtract 4 from 16 to get 12.

$$12-20x-8x^{2}=4\left(2-5x\right)$$

Use the distributive property to multiply 4 by 2-5x.

$$12-20x-8x^{2}=8-20x$$

Subtract 8 from both sides.

$$12-20x-8x^{2}-8=-20x$$

Subtract 8 from 12 to get 4.

$$4-20x-8x^{2}=-20x$$

Add 20x to both sides.

$$4-20x-8x^{2}+20x=0$$

Combine -20x and 20x to get 0.

$$4-8x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-8x^{2}+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 0 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-8\right)\times 4}}{2\left(-8\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-8\right)\times 4}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{0±\sqrt{32\times 4}}{2\left(-8\right)}$$

Multiply 32 times 4.

$$x=\frac{0±\sqrt{128}}{2\left(-8\right)}$$

Take the square root of 128.

$$x=\frac{0±8\sqrt{2}}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{0±8\sqrt{2}}{-16}$$

Now solve the equation x=\frac{0±8\sqrt{2}}{-16} when ± is plus.

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

Now solve the equation x=\frac{0±8\sqrt{2}}{-16} when ± is minus.

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

The equation is now solved.

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