Solve 0=-16x^2+250x+150

$$0 = -16 { x }^{ 2 } +250x+150$$

Solve for x

$$x = \frac{5 \sqrt{721} + 125}{16} \approx 16.203575989$$

$$x=\frac{125-5\sqrt{721}}{16}\approx -0.578575989$$

Steps Using the Quadratic Formula

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

$$-16x^{2}+250x+150=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 250 for b, and 150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-250±\sqrt{250^{2}-4\left(-16\right)\times 150}}{2\left(-16\right)}$$

Square 250.

$$x=\frac{-250±\sqrt{62500-4\left(-16\right)\times 150}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-250±\sqrt{62500+64\times 150}}{2\left(-16\right)}$$

Multiply 64 times 150.

$$x=\frac{-250±\sqrt{62500+9600}}{2\left(-16\right)}$$

Add 62500 to 9600.

$$x=\frac{-250±\sqrt{72100}}{2\left(-16\right)}$$

Take the square root of 72100.

$$x=\frac{-250±10\sqrt{721}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-250±10\sqrt{721}}{-32}$$

Now solve the equation x=\frac{-250±10\sqrt{721}}{-32} when ± is plus. Add -250 to 10\sqrt{721}.

$$x=\frac{10\sqrt{721}-250}{-32}$$

Divide -250+10\sqrt{721} by -32.

$$x=\frac{125-5\sqrt{721}}{16}$$

Now solve the equation x=\frac{-250±10\sqrt{721}}{-32} when ± is minus. Subtract 10\sqrt{721} from -250.

$$x=\frac{-10\sqrt{721}-250}{-32}$$

Divide -250-10\sqrt{721} by -32.

$$x=\frac{5\sqrt{721}+125}{16}$$

The equation is now solved.

$$x=\frac{125-5\sqrt{721}}{16} x=\frac{5\sqrt{721}+125}{16}$$

Steps for Completing the Square

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

$$-16x^{2}+250x+150=0$$

Subtract 150 from both sides. Anything subtracted from zero gives its negation.

$$-16x^{2}+250x=-150$$

Divide both sides by -16.

$$\frac{-16x^{2}+250x}{-16}=-\frac{150}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$x^{2}+\frac{250}{-16}x=-\frac{150}{-16}$$

Reduce the fraction \frac{250}{-16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{125}{8}x=-\frac{150}{-16}$$

Reduce the fraction \frac{-150}{-16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{125}{8}x=\frac{75}{8}$$

Divide -\frac{125}{8}, the coefficient of the x term, by 2 to get -\frac{125}{16}. Then add the square of -\frac{125}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{125}{8}x+\left(-\frac{125}{16}\right)^{2}=\frac{75}{8}+\left(-\frac{125}{16}\right)^{2}$$

Square -\frac{125}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{125}{8}x+\frac{15625}{256}=\frac{75}{8}+\frac{15625}{256}$$

Add \frac{75}{8} to \frac{15625}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{125}{8}x+\frac{15625}{256}=\frac{18025}{256}$$

Factor x^{2}-\frac{125}{8}x+\frac{15625}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{125}{16}\right)^{2}=\frac{18025}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{125}{16}\right)^{2}}=\sqrt{\frac{18025}{256}}$$

Simplify.

$$x-\frac{125}{16}=\frac{5\sqrt{721}}{16} x-\frac{125}{16}=-\frac{5\sqrt{721}}{16}$$

Add \frac{125}{16} to both sides of the equation.

$$x=\frac{5\sqrt{721}+125}{16} x=\frac{125-5\sqrt{721}}{16}$$