Solve 80=-10x^2+80x

$$80=-10 { x }^{ 2 } +80x$$

Solve for x

$$x=2\sqrt{2}+4\approx 6.828427125$$

$$x=4-2\sqrt{2}\approx 1.171572875$$

Steps Using the Quadratic Formula

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

$$-10x^{2}+80x=80$$

Subtract 80 from both sides.

$$-10x^{2}+80x-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 80 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-80±\sqrt{80^{2}-4\left(-10\right)\left(-80\right)}}{2\left(-10\right)}$$

Square 80.

$$x=\frac{-80±\sqrt{6400-4\left(-10\right)\left(-80\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-80±\sqrt{6400+40\left(-80\right)}}{2\left(-10\right)}$$

Multiply 40 times -80.

$$x=\frac{-80±\sqrt{6400-3200}}{2\left(-10\right)}$$

Add 6400 to -3200.

$$x=\frac{-80±\sqrt{3200}}{2\left(-10\right)}$$

Take the square root of 3200.

$$x=\frac{-80±40\sqrt{2}}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-80±40\sqrt{2}}{-20}$$

Now solve the equation x=\frac{-80±40\sqrt{2}}{-20} when ± is plus. Add -80 to 40\sqrt{2}.

$$x=\frac{40\sqrt{2}-80}{-20}$$

Divide -80+40\sqrt{2} by -20.

$$x=4-2\sqrt{2}$$

Now solve the equation x=\frac{-80±40\sqrt{2}}{-20} when ± is minus. Subtract 40\sqrt{2} from -80.

$$x=\frac{-40\sqrt{2}-80}{-20}$$

Divide -80-40\sqrt{2} by -20.

$$x=2\sqrt{2}+4$$

The equation is now solved.

$$x=4-2\sqrt{2} x=2\sqrt{2}+4$$

Steps for Completing the Square

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

$$-10x^{2}+80x=80$$

Divide both sides by -10.

$$\frac{-10x^{2}+80x}{-10}=\frac{80}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{80}{-10}x=\frac{80}{-10}$$

Divide 80 by -10.

$$x^{2}-8x=\frac{80}{-10}$$

Divide 80 by -10.

$$x^{2}-8x=-8$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -4.

$$x^{2}-8x+16=-8+16$$

Add -8 to 16.

$$x^{2}-8x+16=8$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=2\sqrt{2} x-4=-2\sqrt{2}$$

Add 4 to both sides of the equation.

$$x=2\sqrt{2}+4 x=4-2\sqrt{2}$$