Solve 0=-16t^2+88t+4

$$0=-16 { t }^{ 2 } +88t+4$$

Solve for t

$$t = \frac{5 \sqrt{5} + 11}{4} \approx 5.545084972$$

$$t=\frac{11-5\sqrt{5}}{4}\approx -0.045084972$$

Steps Using the Quadratic Formula

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

$$-16t^{2}+88t+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 88 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-88±\sqrt{88^{2}-4\left(-16\right)\times 4}}{2\left(-16\right)}$$

Square 88.

$$t=\frac{-88±\sqrt{7744-4\left(-16\right)\times 4}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-88±\sqrt{7744+64\times 4}}{2\left(-16\right)}$$

Multiply 64 times 4.

$$t=\frac{-88±\sqrt{7744+256}}{2\left(-16\right)}$$

Add 7744 to 256.

$$t=\frac{-88±\sqrt{8000}}{2\left(-16\right)}$$

Take the square root of 8000.

$$t=\frac{-88±40\sqrt{5}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-88±40\sqrt{5}}{-32}$$

Now solve the equation t=\frac{-88±40\sqrt{5}}{-32} when ± is plus. Add -88 to 40\sqrt{5}.

$$t=\frac{40\sqrt{5}-88}{-32}$$

Divide -88+40\sqrt{5} by -32.

$$t=\frac{11-5\sqrt{5}}{4}$$

Now solve the equation t=\frac{-88±40\sqrt{5}}{-32} when ± is minus. Subtract 40\sqrt{5} from -88.

$$t=\frac{-40\sqrt{5}-88}{-32}$$

Divide -88-40\sqrt{5} by -32.

$$t=\frac{5\sqrt{5}+11}{4}$$

The equation is now solved.

$$t=\frac{11-5\sqrt{5}}{4} t=\frac{5\sqrt{5}+11}{4}$$

Steps for Completing the Square

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

$$-16t^{2}+88t+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-16t^{2}+88t=-4$$

Divide both sides by -16.

$$\frac{-16t^{2}+88t}{-16}=-\frac{4}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{88}{-16}t=-\frac{4}{-16}$$

Reduce the fraction \frac{88}{-16} to lowest terms by extracting and canceling out 8.

$$t^{2}-\frac{11}{2}t=-\frac{4}{-16}$$

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

$$t^{2}-\frac{11}{2}t=\frac{1}{4}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{11}{2}t+\left(-\frac{11}{4}\right)^{2}=\frac{1}{4}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{11}{2}t+\frac{121}{16}=\frac{1}{4}+\frac{121}{16}$$

Add \frac{1}{4} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{11}{2}t+\frac{121}{16}=\frac{125}{16}$$

Factor t^{2}-\frac{11}{2}t+\frac{121}{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}.

$$\left(t-\frac{11}{4}\right)^{2}=\frac{125}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{11}{4}\right)^{2}}=\sqrt{\frac{125}{16}}$$

Simplify.

$$t-\frac{11}{4}=\frac{5\sqrt{5}}{4} t-\frac{11}{4}=-\frac{5\sqrt{5}}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$t=\frac{5\sqrt{5}+11}{4} t=\frac{11-5\sqrt{5}}{4}$$