Solve 100+400=1600+x^2-80x

$$100+400=1600+ { x }^{ 2 } -80x$$

Solve for x

$$x=10\sqrt{5}+40\approx 62.360679775$$

$$x=40-10\sqrt{5}\approx 17.639320225$$

Steps Using the Quadratic Formula

Add 100 and 400 to get 500.

$$500=1600+x^{2}-80x$$

Swap sides so that all variable terms are on the left hand side.

$$1600+x^{2}-80x=500$$

Subtract 500 from both sides.

$$1600+x^{2}-80x-500=0$$

Subtract 500 from 1600 to get 1100.

$$1100+x^{2}-80x=0$$

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.

$$x^{2}-80x+1100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -80 for b, and 1100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 1100}}{2}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 1100}}{2}$$

Multiply -4 times 1100.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4400}}{2}$$

Add 6400 to -4400.

$$x=\frac{-\left(-80\right)±\sqrt{2000}}{2}$$

Take the square root of 2000.

$$x=\frac{-\left(-80\right)±20\sqrt{5}}{2}$$

The opposite of -80 is 80.

$$x=\frac{80±20\sqrt{5}}{2}$$

Now solve the equation x=\frac{80±20\sqrt{5}}{2} when ± is plus. Add 80 to 20\sqrt{5}.

$$x=\frac{20\sqrt{5}+80}{2}$$

Divide 80+20\sqrt{5} by 2.

$$x=10\sqrt{5}+40$$

Now solve the equation x=\frac{80±20\sqrt{5}}{2} when ± is minus. Subtract 20\sqrt{5} from 80.

$$x=\frac{80-20\sqrt{5}}{2}$$

Divide 80-20\sqrt{5} by 2.

$$x=40-10\sqrt{5}$$

The equation is now solved.

$$x=10\sqrt{5}+40 x=40-10\sqrt{5}$$

Steps for Completing the Square

Add 100 and 400 to get 500.

$$500=1600+x^{2}-80x$$

Swap sides so that all variable terms are on the left hand side.

$$1600+x^{2}-80x=500$$

Subtract 1600 from both sides.

$$x^{2}-80x=500-1600$$

Subtract 1600 from 500 to get -1100.

$$x^{2}-80x=-1100$$

Divide -80, the coefficient of the x term, by 2 to get -40. Then add the square of -40 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-80x+\left(-40\right)^{2}=-1100+\left(-40\right)^{2}$$

Square -40.

$$x^{2}-80x+1600=-1100+1600$$

Add -1100 to 1600.

$$x^{2}-80x+1600=500$$

Factor x^{2}-80x+1600. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-40\right)^{2}=500$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-40\right)^{2}}=\sqrt{500}$$

Simplify.

$$x-40=10\sqrt{5} x-40=-10\sqrt{5}$$

Add 40 to both sides of the equation.

$$x=10\sqrt{5}+40 x=40-10\sqrt{5}$$