Solve 15t=1/2*10t^2

$$15 t = \frac { 1 } { 2 } \times 10 t ^ { 2 }$$

Solve for t

$$t=3$$

$$t=0$$

Steps Using Factoring

Multiply \frac{1}{2} and 10 to get 5.

$$15t=5t^{2}$$

Subtract 5t^{2} from both sides.

$$15t-5t^{2}=0$$

Factor out t.

$$t\left(15-5t\right)=0$$

To find equation solutions, solve t=0 and 15-5t=0.

$$t=0 t=3$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 10 to get 5.

$$15t=5t^{2}$$

Subtract 5t^{2} from both sides.

$$15t-5t^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-5t^{2}+15t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 15 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-15±\sqrt{15^{2}}}{2\left(-5\right)}$$

Take the square root of 15^{2}.

$$t=\frac{-15±15}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-15±15}{-10}$$

Now solve the equation t=\frac{-15±15}{-10} when ± is plus. Add -15 to 15.

$$t=\frac{0}{-10}$$

Divide 0 by -10.

$$t=0$$

Now solve the equation t=\frac{-15±15}{-10} when ± is minus. Subtract 15 from -15.

$$t=-\frac{30}{-10}$$

Divide -30 by -10.

$$t=3$$

The equation is now solved.

$$t=0 t=3$$

Steps for Completing the Square

Multiply \frac{1}{2} and 10 to get 5.

$$15t=5t^{2}$$

Subtract 5t^{2} from both sides.

$$15t-5t^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5t^{2}+15t=0$$

Divide both sides by -5.

$$\frac{-5t^{2}+15t}{-5}=\frac{0}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\frac{15}{-5}t=\frac{0}{-5}$$

Divide 15 by -5.

$$t^{2}-3t=\frac{0}{-5}$$

Divide 0 by -5.

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

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.

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

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

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

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}.

$$\left(t-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

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

$$t=3 t=0$$