Solve 420=-16t^2+180t

$$420 = - 16 t ^ { 2 } + 180 t$$

Solve for t

$$t = \frac{\sqrt{345} + 45}{8} \approx 7.946771953$$

$$t = \frac{45 - \sqrt{345}}{8} \approx 3.303228047$$

Steps Using the Quadratic Formula

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

$$-16t^{2}+180t=420$$

Subtract 420 from both sides.

$$-16t^{2}+180t-420=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 180 for b, and -420 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-180±\sqrt{180^{2}-4\left(-16\right)\left(-420\right)}}{2\left(-16\right)}$$

Square 180.

$$t=\frac{-180±\sqrt{32400-4\left(-16\right)\left(-420\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-180±\sqrt{32400+64\left(-420\right)}}{2\left(-16\right)}$$

Multiply 64 times -420.

$$t=\frac{-180±\sqrt{32400-26880}}{2\left(-16\right)}$$

Add 32400 to -26880.

$$t=\frac{-180±\sqrt{5520}}{2\left(-16\right)}$$

Take the square root of 5520.

$$t=\frac{-180±4\sqrt{345}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-180±4\sqrt{345}}{-32}$$

Now solve the equation t=\frac{-180±4\sqrt{345}}{-32} when ± is plus. Add -180 to 4\sqrt{345}.

$$t=\frac{4\sqrt{345}-180}{-32}$$

Divide -180+4\sqrt{345} by -32.

$$t=\frac{45-\sqrt{345}}{8}$$

Now solve the equation t=\frac{-180±4\sqrt{345}}{-32} when ± is minus. Subtract 4\sqrt{345} from -180.

$$t=\frac{-4\sqrt{345}-180}{-32}$$

Divide -180-4\sqrt{345} by -32.

$$t=\frac{\sqrt{345}+45}{8}$$

The equation is now solved.

$$t=\frac{45-\sqrt{345}}{8} t=\frac{\sqrt{345}+45}{8}$$

Steps for Completing the Square

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

$$-16t^{2}+180t=420$$

Divide both sides by -16.

$$\frac{-16t^{2}+180t}{-16}=\frac{420}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{180}{-16}t=\frac{420}{-16}$$

Reduce the fraction \frac{180}{-16} to lowest terms by extracting and canceling out 4.

$$t^{2}-\frac{45}{4}t=\frac{420}{-16}$$

Reduce the fraction \frac{420}{-16} to lowest terms by extracting and canceling out 4.

$$t^{2}-\frac{45}{4}t=-\frac{105}{4}$$

Divide -\frac{45}{4}, the coefficient of the x term, by 2 to get -\frac{45}{8}. Then add the square of -\frac{45}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{45}{4}t+\left(-\frac{45}{8}\right)^{2}=-\frac{105}{4}+\left(-\frac{45}{8}\right)^{2}$$

Square -\frac{45}{8} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{45}{4}t+\frac{2025}{64}=-\frac{105}{4}+\frac{2025}{64}$$

Add -\frac{105}{4} to \frac{2025}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{45}{4}t+\frac{2025}{64}=\frac{345}{64}$$

Factor t^{2}-\frac{45}{4}t+\frac{2025}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{45}{8}\right)^{2}=\frac{345}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{45}{8}\right)^{2}}=\sqrt{\frac{345}{64}}$$

Simplify.

$$t-\frac{45}{8}=\frac{\sqrt{345}}{8} t-\frac{45}{8}=-\frac{\sqrt{345}}{8}$$

Add \frac{45}{8} to both sides of the equation.

$$t=\frac{\sqrt{345}+45}{8} t=\frac{45-\sqrt{345}}{8}$$