Solve 15t-3/2t^2=36t-9t^2

$$15 t - \frac { 3 } { 2 } t ^ { 2 } = 36 t - 9 t ^ { 2 }$$

Solve for t

$$t = \frac{14}{5} = 2\frac{4}{5} = 2.8$$

$$t=0$$

Steps Using Factoring

Subtract 36t from both sides.

$$15t-\frac{3}{2}t^{2}-36t=-9t^{2}$$

Combine 15t and -36t to get -21t.

$$-21t-\frac{3}{2}t^{2}=-9t^{2}$$

Add 9t^{2} to both sides.

$$-21t-\frac{3}{2}t^{2}+9t^{2}=0$$

Combine -\frac{3}{2}t^{2} and 9t^{2} to get \frac{15}{2}t^{2}.

$$-21t+\frac{15}{2}t^{2}=0$$

Factor out t.

$$t\left(-21+\frac{15}{2}t\right)=0$$

To find equation solutions, solve t=0 and -21+\frac{15t}{2}=0.

$$t=0 t=\frac{14}{5}$$

Steps Using the Quadratic Formula

Subtract 36t from both sides.

$$15t-\frac{3}{2}t^{2}-36t=-9t^{2}$$

Combine 15t and -36t to get -21t.

$$-21t-\frac{3}{2}t^{2}=-9t^{2}$$

Add 9t^{2} to both sides.

$$-21t-\frac{3}{2}t^{2}+9t^{2}=0$$

Combine -\frac{3}{2}t^{2} and 9t^{2} to get \frac{15}{2}t^{2}.

$$-21t+\frac{15}{2}t^{2}=0$$

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.

$$\frac{15}{2}t^{2}-21t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{15}{2} for a, -21 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}}}{2\times \frac{15}{2}}$$

Take the square root of \left(-21\right)^{2}.

$$t=\frac{-\left(-21\right)±21}{2\times \frac{15}{2}}$$

The opposite of -21 is 21.

$$t=\frac{21±21}{2\times \frac{15}{2}}$$

Multiply 2 times \frac{15}{2}.

$$t=\frac{21±21}{15}$$

Now solve the equation t=\frac{21±21}{15} when ± is plus. Add 21 to 21.

$$t=\frac{42}{15}$$

Reduce the fraction \frac{42}{15} to lowest terms by extracting and canceling out 3.

$$t=\frac{14}{5}$$

Now solve the equation t=\frac{21±21}{15} when ± is minus. Subtract 21 from 21.

$$t=\frac{0}{15}$$

Divide 0 by 15.

$$t=0$$

The equation is now solved.

$$t=\frac{14}{5} t=0$$

Steps for Completing the Square

Subtract 36t from both sides.

$$15t-\frac{3}{2}t^{2}-36t=-9t^{2}$$

Combine 15t and -36t to get -21t.

$$-21t-\frac{3}{2}t^{2}=-9t^{2}$$

Add 9t^{2} to both sides.

$$-21t-\frac{3}{2}t^{2}+9t^{2}=0$$

Combine -\frac{3}{2}t^{2} and 9t^{2} to get \frac{15}{2}t^{2}.

$$-21t+\frac{15}{2}t^{2}=0$$

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.

$$\frac{15}{2}t^{2}-21t=0$$

Divide both sides of the equation by \frac{15}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{15}{2}t^{2}-21t}{\frac{15}{2}}=\frac{0}{\frac{15}{2}}$$

Dividing by \frac{15}{2} undoes the multiplication by \frac{15}{2}.

$$t^{2}+\left(-\frac{21}{\frac{15}{2}}\right)t=\frac{0}{\frac{15}{2}}$$

Divide -21 by \frac{15}{2} by multiplying -21 by the reciprocal of \frac{15}{2}.

$$t^{2}-\frac{14}{5}t=\frac{0}{\frac{15}{2}}$$

Divide 0 by \frac{15}{2} by multiplying 0 by the reciprocal of \frac{15}{2}.

$$t^{2}-\frac{14}{5}t=0$$

Divide -\frac{14}{5}, the coefficient of the x term, by 2 to get -\frac{7}{5}. Then add the square of -\frac{7}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{14}{5}t+\left(-\frac{7}{5}\right)^{2}=\left(-\frac{7}{5}\right)^{2}$$

Square -\frac{7}{5} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{14}{5}t+\frac{49}{25}=\frac{49}{25}$$

Factor t^{2}-\frac{14}{5}t+\frac{49}{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}.

$$\left(t-\frac{7}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{7}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

$$t-\frac{7}{5}=\frac{7}{5} t-\frac{7}{5}=-\frac{7}{5}$$

Add \frac{7}{5} to both sides of the equation.

$$t=\frac{14}{5} t=0$$