Solve (1/2)32t^2+16t=480

$$( 1 / 2 ) 32 t ^ { 2 } + 16 t = 480$$

Solve for t

$$t=-6$$

$$t=5$$

Steps Using Factoring By Grouping

Multiply \frac{1}{2} and 32 to get 16.

$$16t^{2}+16t=480$$

Subtract 480 from both sides.

$$16t^{2}+16t-480=0$$

Divide both sides by 16.

$$t^{2}+t-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-30\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite t^{2}+t-30 as \left(t^{2}-5t\right)+\left(6t-30\right).

$$\left(t^{2}-5t\right)+\left(6t-30\right)$$

Factor out t in the first and 6 in the second group.

$$t\left(t-5\right)+6\left(t-5\right)$$

Factor out common term t-5 by using distributive property.

$$\left(t-5\right)\left(t+6\right)$$

To find equation solutions, solve t-5=0 and t+6=0.

$$t=5 t=-6$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 32 to get 16.

$$16t^{2}+16t=480$$

Subtract 480 from both sides.

$$16t^{2}+16t-480=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 16 for b, and -480 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-16±\sqrt{16^{2}-4\times 16\left(-480\right)}}{2\times 16}$$

Square 16.

$$t=\frac{-16±\sqrt{256-4\times 16\left(-480\right)}}{2\times 16}$$

Multiply -4 times 16.

$$t=\frac{-16±\sqrt{256-64\left(-480\right)}}{2\times 16}$$

Multiply -64 times -480.

$$t=\frac{-16±\sqrt{256+30720}}{2\times 16}$$

Add 256 to 30720.

$$t=\frac{-16±\sqrt{30976}}{2\times 16}$$

Take the square root of 30976.

$$t=\frac{-16±176}{2\times 16}$$

Multiply 2 times 16.

$$t=\frac{-16±176}{32}$$

Now solve the equation t=\frac{-16±176}{32} when ± is plus. Add -16 to 176.

$$t=\frac{160}{32}$$

Divide 160 by 32.

$$t=5$$

Now solve the equation t=\frac{-16±176}{32} when ± is minus. Subtract 176 from -16.

$$t=-\frac{192}{32}$$

Divide -192 by 32.

$$t=-6$$

The equation is now solved.

$$t=5 t=-6$$

Steps for Completing the Square

Multiply \frac{1}{2} and 32 to get 16.

$$16t^{2}+16t=480$$

Divide both sides by 16.

$$\frac{16t^{2}+16t}{16}=\frac{480}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$t^{2}+\frac{16}{16}t=\frac{480}{16}$$

Divide 16 by 16.

$$t^{2}+t=\frac{480}{16}$$

Divide 480 by 16.

$$t^{2}+t=30$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+t+\left(\frac{1}{2}\right)^{2}=30+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+t+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

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

Factor t^{2}+t+\frac{1}{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}.

$$\left(t+\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$t+\frac{1}{2}=\frac{11}{2} t+\frac{1}{2}=-\frac{11}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$t=5 t=-6$$