Solve 4x^2=16left(120-xright)^2

$$4 { x }^{ 2 } =16 { \left(120-x \right) }^{ 2 }$$

Solve for x

$$x=240$$

$$x=80$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(120-x\right)^{2}.

$$4x^{2}=16\left(14400-240x+x^{2}\right)$$

Use the distributive property to multiply 16 by 14400-240x+x^{2}.

$$4x^{2}=230400-3840x+16x^{2}$$

Subtract 230400 from both sides.

$$4x^{2}-230400=-3840x+16x^{2}$$

Add 3840x to both sides.

$$4x^{2}-230400+3840x=16x^{2}$$

Subtract 16x^{2} from both sides.

$$4x^{2}-230400+3840x-16x^{2}=0$$

Combine 4x^{2} and -16x^{2} to get -12x^{2}.

$$-12x^{2}-230400+3840x=0$$

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.

$$-12x^{2}+3840x-230400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 3840 for b, and -230400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3840±\sqrt{3840^{2}-4\left(-12\right)\left(-230400\right)}}{2\left(-12\right)}$$

Square 3840.

$$x=\frac{-3840±\sqrt{14745600-4\left(-12\right)\left(-230400\right)}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-3840±\sqrt{14745600+48\left(-230400\right)}}{2\left(-12\right)}$$

Multiply 48 times -230400.

$$x=\frac{-3840±\sqrt{14745600-11059200}}{2\left(-12\right)}$$

Add 14745600 to -11059200.

$$x=\frac{-3840±\sqrt{3686400}}{2\left(-12\right)}$$

Take the square root of 3686400.

$$x=\frac{-3840±1920}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-3840±1920}{-24}$$

Now solve the equation x=\frac{-3840±1920}{-24} when ± is plus. Add -3840 to 1920.

$$x=-\frac{1920}{-24}$$

Divide -1920 by -24.

$$x=80$$

Now solve the equation x=\frac{-3840±1920}{-24} when ± is minus. Subtract 1920 from -3840.

$$x=-\frac{5760}{-24}$$

Divide -5760 by -24.

$$x=240$$

The equation is now solved.

$$x=80 x=240$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(120-x\right)^{2}.

$$4x^{2}=16\left(14400-240x+x^{2}\right)$$

Use the distributive property to multiply 16 by 14400-240x+x^{2}.

$$4x^{2}=230400-3840x+16x^{2}$$

Add 3840x to both sides.

$$4x^{2}+3840x=230400+16x^{2}$$

Subtract 16x^{2} from both sides.

$$4x^{2}+3840x-16x^{2}=230400$$

Combine 4x^{2} and -16x^{2} to get -12x^{2}.

$$-12x^{2}+3840x=230400$$

Divide both sides by -12.

$$\frac{-12x^{2}+3840x}{-12}=\frac{230400}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{3840}{-12}x=\frac{230400}{-12}$$

Divide 3840 by -12.

$$x^{2}-320x=\frac{230400}{-12}$$

Divide 230400 by -12.

$$x^{2}-320x=-19200$$

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.

$$x^{2}-320x+\left(-160\right)^{2}=-19200+\left(-160\right)^{2}$$

Square -160.

$$x^{2}-320x+25600=-19200+25600$$

Add -19200 to 25600.

$$x^{2}-320x+25600=6400$$

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-160\right)^{2}}=\sqrt{6400}$$

Simplify.

$$x-160=80 x-160=-80$$

Add 160 to both sides of the equation.

$$x=240 x=80$$