Solve 0=-16t^2+48t-32

$$0 = - 16 t ^ { 2 } + 48 t - 32$$

Solve for t

$$t=1$$

$$t=2$$

Steps Using Factoring By Grouping

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

$$-16t^{2}+48t-32=0$$

Divide both sides by 16.

$$-t^{2}+3t-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt-2. To find a and b, set up a system to be solved.

$$a+b=3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=2 b=1$$

Rewrite -t^{2}+3t-2 as \left(-t^{2}+2t\right)+\left(t-2\right).

$$\left(-t^{2}+2t\right)+\left(t-2\right)$$

Factor out -t in -t^{2}+2t.

$$-t\left(t-2\right)+t-2$$

Factor out common term t-2 by using distributive property.

$$\left(t-2\right)\left(-t+1\right)$$

To find equation solutions, solve t-2=0 and -t+1=0.

$$t=2 t=1$$

Steps Using the Quadratic Formula

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

$$-16t^{2}+48t-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 48 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-48±\sqrt{48^{2}-4\left(-16\right)\left(-32\right)}}{2\left(-16\right)}$$

Square 48.

$$t=\frac{-48±\sqrt{2304-4\left(-16\right)\left(-32\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-48±\sqrt{2304+64\left(-32\right)}}{2\left(-16\right)}$$

Multiply 64 times -32.

$$t=\frac{-48±\sqrt{2304-2048}}{2\left(-16\right)}$$

Add 2304 to -2048.

$$t=\frac{-48±\sqrt{256}}{2\left(-16\right)}$$

Take the square root of 256.

$$t=\frac{-48±16}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-48±16}{-32}$$

Now solve the equation t=\frac{-48±16}{-32} when ± is plus. Add -48 to 16.

$$t=-\frac{32}{-32}$$

Divide -32 by -32.

$$t=1$$

Now solve the equation t=\frac{-48±16}{-32} when ± is minus. Subtract 16 from -48.

$$t=-\frac{64}{-32}$$

Divide -64 by -32.

$$t=2$$

The equation is now solved.

$$t=1 t=2$$

Steps for Completing the Square

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

$$-16t^{2}+48t-32=0$$

Add 32 to both sides. Anything plus zero gives itself.

$$-16t^{2}+48t=32$$

Divide both sides by -16.

$$\frac{-16t^{2}+48t}{-16}=\frac{32}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{48}{-16}t=\frac{32}{-16}$$

Divide 48 by -16.

$$t^{2}-3t=\frac{32}{-16}$$

Divide 32 by -16.

$$t^{2}-3t=-2$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-3t+\left(-\frac{3}{2}\right)^{2}=-2+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-3t+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$t^{2}-3t+\frac{9}{4}=\frac{1}{4}$$

Factor t^{2}-3t+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$t-\frac{3}{2}=\frac{1}{2} t-\frac{3}{2}=-\frac{1}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$t=2 t=1$$