Solve 4x^2=8left(80-xright)^2

$$4 { x }^{ 2 } =8 { \left(80-x \right) }^{ 2 }$$

Solve for x

$$x=80\sqrt{2}+160\approx 273.13708499$$

$$x=160-80\sqrt{2}\approx 46.86291501$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(80-x\right)^{2}.

$$4x^{2}=8\left(6400-160x+x^{2}\right)$$

Use the distributive property to multiply 8 by 6400-160x+x^{2}.

$$4x^{2}=51200-1280x+8x^{2}$$

Subtract 51200 from both sides.

$$4x^{2}-51200=-1280x+8x^{2}$$

Add 1280x to both sides.

$$4x^{2}-51200+1280x=8x^{2}$$

Subtract 8x^{2} from both sides.

$$4x^{2}-51200+1280x-8x^{2}=0$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$-4x^{2}-51200+1280x=0$$

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.

$$-4x^{2}+1280x-51200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 1280 for b, and -51200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1280±\sqrt{1280^{2}-4\left(-4\right)\left(-51200\right)}}{2\left(-4\right)}$$

Square 1280.

$$x=\frac{-1280±\sqrt{1638400-4\left(-4\right)\left(-51200\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-1280±\sqrt{1638400+16\left(-51200\right)}}{2\left(-4\right)}$$

Multiply 16 times -51200.

$$x=\frac{-1280±\sqrt{1638400-819200}}{2\left(-4\right)}$$

Add 1638400 to -819200.

$$x=\frac{-1280±\sqrt{819200}}{2\left(-4\right)}$$

Take the square root of 819200.

$$x=\frac{-1280±640\sqrt{2}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-1280±640\sqrt{2}}{-8}$$

Now solve the equation x=\frac{-1280±640\sqrt{2}}{-8} when ± is plus. Add -1280 to 640\sqrt{2}.

$$x=\frac{640\sqrt{2}-1280}{-8}$$

Divide -1280+640\sqrt{2} by -8.

$$x=160-80\sqrt{2}$$

Now solve the equation x=\frac{-1280±640\sqrt{2}}{-8} when ± is minus. Subtract 640\sqrt{2} from -1280.

$$x=\frac{-640\sqrt{2}-1280}{-8}$$

Divide -1280-640\sqrt{2} by -8.

$$x=80\sqrt{2}+160$$

The equation is now solved.

$$x=160-80\sqrt{2} x=80\sqrt{2}+160$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(80-x\right)^{2}.

$$4x^{2}=8\left(6400-160x+x^{2}\right)$$

Use the distributive property to multiply 8 by 6400-160x+x^{2}.

$$4x^{2}=51200-1280x+8x^{2}$$

Add 1280x to both sides.

$$4x^{2}+1280x=51200+8x^{2}$$

Subtract 8x^{2} from both sides.

$$4x^{2}+1280x-8x^{2}=51200$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$-4x^{2}+1280x=51200$$

Divide both sides by -4.

$$\frac{-4x^{2}+1280x}{-4}=\frac{51200}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{1280}{-4}x=\frac{51200}{-4}$$

Divide 1280 by -4.

$$x^{2}-320x=\frac{51200}{-4}$$

Divide 51200 by -4.

$$x^{2}-320x=-12800$$

Divide -320, the coefficient of the x term, by 2 to get -160. Then add the square of -160 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-320x+\left(-160\right)^{2}=-12800+\left(-160\right)^{2}$$

Square -160.

$$x^{2}-320x+25600=-12800+25600$$

Add -12800 to 25600.

$$x^{2}-320x+25600=12800$$

Factor x^{2}-320x+25600. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-160\right)^{2}=12800$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-160\right)^{2}}=\sqrt{12800}$$

Simplify.

$$x-160=80\sqrt{2} x-160=-80\sqrt{2}$$

Add 160 to both sides of the equation.

$$x=80\sqrt{2}+160 x=160-80\sqrt{2}$$