Solve -20x^2+520x-3200=0

$$-20 { x }^{ 2 } +520x-3200=0$$

Solve for x

$$x=10$$

$$x=16$$

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.

$$-20x^{2}+520x-3200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -20 for a, 520 for b, and -3200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-520±\sqrt{520^{2}-4\left(-20\right)\left(-3200\right)}}{2\left(-20\right)}$$

Square 520.

$$x=\frac{-520±\sqrt{270400-4\left(-20\right)\left(-3200\right)}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-520±\sqrt{270400+80\left(-3200\right)}}{2\left(-20\right)}$$

Multiply 80 times -3200.

$$x=\frac{-520±\sqrt{270400-256000}}{2\left(-20\right)}$$

Add 270400 to -256000.

$$x=\frac{-520±\sqrt{14400}}{2\left(-20\right)}$$

Take the square root of 14400.

$$x=\frac{-520±120}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-520±120}{-40}$$

Now solve the equation x=\frac{-520±120}{-40} when ± is plus. Add -520 to 120.

$$x=-\frac{400}{-40}$$

Divide -400 by -40.

$$x=10$$

Now solve the equation x=\frac{-520±120}{-40} when ± is minus. Subtract 120 from -520.

$$x=-\frac{640}{-40}$$

Divide -640 by -40.

$$x=16$$

The equation is now solved.

$$x=10 x=16$$

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.

$$-20x^{2}+520x-3200=0$$

Add 3200 to both sides of the equation.

$$-20x^{2}+520x-3200-\left(-3200\right)=-\left(-3200\right)$$

Subtracting -3200 from itself leaves 0.

$$-20x^{2}+520x=-\left(-3200\right)$$

Subtract -3200 from 0.

$$-20x^{2}+520x=3200$$

Divide both sides by -20.

$$\frac{-20x^{2}+520x}{-20}=\frac{3200}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$x^{2}+\frac{520}{-20}x=\frac{3200}{-20}$$

Divide 520 by -20.

$$x^{2}-26x=\frac{3200}{-20}$$

Divide 3200 by -20.

$$x^{2}-26x=-160$$

Divide -26, the coefficient of the x term, by 2 to get -13. Then add the square of -13 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-26x+\left(-13\right)^{2}=-160+\left(-13\right)^{2}$$

Square -13.

$$x^{2}-26x+169=-160+169$$

Add -160 to 169.

$$x^{2}-26x+169=9$$

Factor x^{2}-26x+169. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-13\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-13\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-13=3 x-13=-3$$

Add 13 to both sides of the equation.

$$x=16 x=10$$