Solve x^2-80x+300=0

$${ x }^{ 2 } -80x+300=0$$

Solve for x

$$x=10\sqrt{13}+40\approx 76.055512755$$

$$x=40-10\sqrt{13}\approx 3.944487245$$

Steps Using the Quadratic Formula

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.

$$x^{2}-80x+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -80 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 300}}{2}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 300}}{2}$$

Multiply -4 times 300.

$$x=\frac{-\left(-80\right)±\sqrt{6400-1200}}{2}$$

Add 6400 to -1200.

$$x=\frac{-\left(-80\right)±\sqrt{5200}}{2}$$

Take the square root of 5200.

$$x=\frac{-\left(-80\right)±20\sqrt{13}}{2}$$

The opposite of -80 is 80.

$$x=\frac{80±20\sqrt{13}}{2}$$

Now solve the equation x=\frac{80±20\sqrt{13}}{2} when ± is plus. Add 80 to 20\sqrt{13}.

$$x=\frac{20\sqrt{13}+80}{2}$$

Divide 80+20\sqrt{13} by 2.

$$x=10\sqrt{13}+40$$

Now solve the equation x=\frac{80±20\sqrt{13}}{2} when ± is minus. Subtract 20\sqrt{13} from 80.

$$x=\frac{80-20\sqrt{13}}{2}$$

Divide 80-20\sqrt{13} by 2.

$$x=40-10\sqrt{13}$$

The equation is now solved.

$$x=10\sqrt{13}+40 x=40-10\sqrt{13}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}-80x+300=0$$

Subtract 300 from both sides of the equation.

$$x^{2}-80x+300-300=-300$$

Subtracting 300 from itself leaves 0.

$$x^{2}-80x=-300$$

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.

$$x^{2}-80x+\left(-40\right)^{2}=-300+\left(-40\right)^{2}$$

Square -40.

$$x^{2}-80x+1600=-300+1600$$

Add -300 to 1600.

$$x^{2}-80x+1600=1300$$

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}.

$$\left(x-40\right)^{2}=1300$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-40\right)^{2}}=\sqrt{1300}$$

Simplify.

$$x-40=10\sqrt{13} x-40=-10\sqrt{13}$$

Add 40 to both sides of the equation.

$$x=10\sqrt{13}+40 x=40-10\sqrt{13}$$