Solve 1/2*3*t^2=15t

$$\frac { 1 } { 2 } \cdot 3 \cdot t ^ { 2 } = 15 t$$

Solve for t

$$t=10$$

$$t=0$$

Steps Using Factoring

Multiply \frac{1}{2} and 3 to get \frac{3}{2}.

$$\frac{3}{2}t^{2}=15t$$

Subtract 15t from both sides.

$$\frac{3}{2}t^{2}-15t=0$$

Factor out t.

$$t\left(\frac{3}{2}t-15\right)=0$$

To find equation solutions, solve t=0 and \frac{3t}{2}-15=0.

$$t=0 t=10$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 3 to get \frac{3}{2}.

$$\frac{3}{2}t^{2}=15t$$

Subtract 15t from both sides.

$$\frac{3}{2}t^{2}-15t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{2} for a, -15 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}}}{2\times \frac{3}{2}}$$

Take the square root of \left(-15\right)^{2}.

$$t=\frac{-\left(-15\right)±15}{2\times \frac{3}{2}}$$

The opposite of -15 is 15.

$$t=\frac{15±15}{2\times \frac{3}{2}}$$

Multiply 2 times \frac{3}{2}.

$$t=\frac{15±15}{3}$$

Now solve the equation t=\frac{15±15}{3} when ± is plus. Add 15 to 15.

$$t=\frac{30}{3}$$

Divide 30 by 3.

$$t=10$$

Now solve the equation t=\frac{15±15}{3} when ± is minus. Subtract 15 from 15.

$$t=\frac{0}{3}$$

Divide 0 by 3.

$$t=0$$

The equation is now solved.

$$t=10 t=0$$

Steps for Completing the Square

Multiply \frac{1}{2} and 3 to get \frac{3}{2}.

$$\frac{3}{2}t^{2}=15t$$

Subtract 15t from both sides.

$$\frac{3}{2}t^{2}-15t=0$$

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{2}t^{2}-15t}{\frac{3}{2}}=\frac{0}{\frac{3}{2}}$$

Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.

$$t^{2}+\left(-\frac{15}{\frac{3}{2}}\right)t=\frac{0}{\frac{3}{2}}$$

Divide -15 by \frac{3}{2} by multiplying -15 by the reciprocal of \frac{3}{2}.

$$t^{2}-10t=\frac{0}{\frac{3}{2}}$$

Divide 0 by \frac{3}{2} by multiplying 0 by the reciprocal of \frac{3}{2}.

$$t^{2}-10t=0$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-10t+\left(-5\right)^{2}=\left(-5\right)^{2}$$

Square -5.

$$t^{2}-10t+25=25$$

Factor t^{2}-10t+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-5\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-5\right)^{2}}=\sqrt{25}$$

Simplify.

$$t-5=5 t-5=-5$$

Add 5 to both sides of the equation.

$$t=10 t=0$$