Solve x^2=100^2+(300-2x)^2

$$x ^ { 2 } = 100 ^ { 2 } + ( 300 - 2 x ) ^ { 2 }$$

Solve for x

$$x=\frac{100\sqrt{6}}{3}+200\approx 281.649658093$$

$$x=-\frac{100\sqrt{6}}{3}+200\approx 118.350341907$$

Steps Using the Quadratic Formula

Calculate 100 to the power of 2 and get 10000.

$$x^{2}=10000+\left(300-2x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(300-2x\right)^{2}.

$$x^{2}=10000+90000-1200x+4x^{2}$$

Add 10000 and 90000 to get 100000.

$$x^{2}=100000-1200x+4x^{2}$$

Subtract 100000 from both sides.

$$x^{2}-100000=-1200x+4x^{2}$$

Add 1200x to both sides.

$$x^{2}-100000+1200x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}-100000+1200x-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-100000+1200x=0$$

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.

$$-3x^{2}+1200x-100000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 1200 for b, and -100000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1200±\sqrt{1200^{2}-4\left(-3\right)\left(-100000\right)}}{2\left(-3\right)}$$

Square 1200.

$$x=\frac{-1200±\sqrt{1440000-4\left(-3\right)\left(-100000\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-1200±\sqrt{1440000+12\left(-100000\right)}}{2\left(-3\right)}$$

Multiply 12 times -100000.

$$x=\frac{-1200±\sqrt{1440000-1200000}}{2\left(-3\right)}$$

Add 1440000 to -1200000.

$$x=\frac{-1200±\sqrt{240000}}{2\left(-3\right)}$$

Take the square root of 240000.

$$x=\frac{-1200±200\sqrt{6}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-1200±200\sqrt{6}}{-6}$$

Now solve the equation x=\frac{-1200±200\sqrt{6}}{-6} when ± is plus. Add -1200 to 200\sqrt{6}.

$$x=\frac{200\sqrt{6}-1200}{-6}$$

Divide -1200+200\sqrt{6} by -6.

$$x=-\frac{100\sqrt{6}}{3}+200$$

Now solve the equation x=\frac{-1200±200\sqrt{6}}{-6} when ± is minus. Subtract 200\sqrt{6} from -1200.

$$x=\frac{-200\sqrt{6}-1200}{-6}$$

Divide -1200-200\sqrt{6} by -6.

$$x=\frac{100\sqrt{6}}{3}+200$$

The equation is now solved.

$$x=-\frac{100\sqrt{6}}{3}+200 x=\frac{100\sqrt{6}}{3}+200$$

Steps for Completing the Square

Calculate 100 to the power of 2 and get 10000.

$$x^{2}=10000+\left(300-2x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(300-2x\right)^{2}.

$$x^{2}=10000+90000-1200x+4x^{2}$$

Add 10000 and 90000 to get 100000.

$$x^{2}=100000-1200x+4x^{2}$$

Add 1200x to both sides.

$$x^{2}+1200x=100000+4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+1200x-4x^{2}=100000$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+1200x=100000$$

Divide both sides by -3.

$$\frac{-3x^{2}+1200x}{-3}=\frac{100000}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{1200}{-3}x=\frac{100000}{-3}$$

Divide 1200 by -3.

$$x^{2}-400x=\frac{100000}{-3}$$

Divide 100000 by -3.

$$x^{2}-400x=-\frac{100000}{3}$$

Divide -400, the coefficient of the x term, by 2 to get -200. Then add the square of -200 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-400x+\left(-200\right)^{2}=-\frac{100000}{3}+\left(-200\right)^{2}$$

Square -200.

$$x^{2}-400x+40000=-\frac{100000}{3}+40000$$

Add -\frac{100000}{3} to 40000.

$$x^{2}-400x+40000=\frac{20000}{3}$$

Factor x^{2}-400x+40000. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-200\right)^{2}=\frac{20000}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-200\right)^{2}}=\sqrt{\frac{20000}{3}}$$

Simplify.

$$x-200=\frac{100\sqrt{6}}{3} x-200=-\frac{100\sqrt{6}}{3}$$

Add 200 to both sides of the equation.

$$x=\frac{100\sqrt{6}}{3}+200 x=-\frac{100\sqrt{6}}{3}+200$$