Solve 48/5=-3/5t^2+6t

$$\frac { 48 } { 5 } = - \frac { 3 } { 5 } t ^ { 2 } + 6 t$$

Solve for t

$$t=2$$

$$t=8$$

Steps Using the Quadratic Formula

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

$$-\frac{3}{5}t^{2}+6t=\frac{48}{5}$$

Subtract \frac{48}{5} from both sides.

$$-\frac{3}{5}t^{2}+6t-\frac{48}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{5} for a, 6 for b, and -\frac{48}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-6±\sqrt{6^{2}-4\left(-\frac{3}{5}\right)\left(-\frac{48}{5}\right)}}{2\left(-\frac{3}{5}\right)}$$

Square 6.

$$t=\frac{-6±\sqrt{36-4\left(-\frac{3}{5}\right)\left(-\frac{48}{5}\right)}}{2\left(-\frac{3}{5}\right)}$$

Multiply -4 times -\frac{3}{5}.

$$t=\frac{-6±\sqrt{36+\frac{12}{5}\left(-\frac{48}{5}\right)}}{2\left(-\frac{3}{5}\right)}$$

Multiply \frac{12}{5} times -\frac{48}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-6±\sqrt{36-\frac{576}{25}}}{2\left(-\frac{3}{5}\right)}$$

Add 36 to -\frac{576}{25}.

$$t=\frac{-6±\sqrt{\frac{324}{25}}}{2\left(-\frac{3}{5}\right)}$$

Take the square root of \frac{324}{25}.

$$t=\frac{-6±\frac{18}{5}}{2\left(-\frac{3}{5}\right)}$$

Multiply 2 times -\frac{3}{5}.

$$t=\frac{-6±\frac{18}{5}}{-\frac{6}{5}}$$

Now solve the equation t=\frac{-6±\frac{18}{5}}{-\frac{6}{5}} when ± is plus. Add -6 to \frac{18}{5}.

$$t=-\frac{\frac{12}{5}}{-\frac{6}{5}}$$

Divide -\frac{12}{5} by -\frac{6}{5} by multiplying -\frac{12}{5} by the reciprocal of -\frac{6}{5}.

$$t=2$$

Now solve the equation t=\frac{-6±\frac{18}{5}}{-\frac{6}{5}} when ± is minus. Subtract \frac{18}{5} from -6.

$$t=-\frac{\frac{48}{5}}{-\frac{6}{5}}$$

Divide -\frac{48}{5} by -\frac{6}{5} by multiplying -\frac{48}{5} by the reciprocal of -\frac{6}{5}.

$$t=8$$

The equation is now solved.

$$t=2 t=8$$

Steps for Completing the Square

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

$$-\frac{3}{5}t^{2}+6t=\frac{48}{5}$$

Divide both sides of the equation by -\frac{3}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{5}t^{2}+6t}{-\frac{3}{5}}=\frac{\frac{48}{5}}{-\frac{3}{5}}$$

Dividing by -\frac{3}{5} undoes the multiplication by -\frac{3}{5}.

$$t^{2}+\frac{6}{-\frac{3}{5}}t=\frac{\frac{48}{5}}{-\frac{3}{5}}$$

Divide 6 by -\frac{3}{5} by multiplying 6 by the reciprocal of -\frac{3}{5}.

$$t^{2}-10t=\frac{\frac{48}{5}}{-\frac{3}{5}}$$

Divide \frac{48}{5} by -\frac{3}{5} by multiplying \frac{48}{5} by the reciprocal of -\frac{3}{5}.

$$t^{2}-10t=-16$$

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.

$$t^{2}-10t+\left(-5\right)^{2}=-16+\left(-5\right)^{2}$$

Square -5.

$$t^{2}-10t+25=-16+25$$

Add -16 to 25.

$$t^{2}-10t+25=9$$

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}.

$$\left(t-5\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-5\right)^{2}}=\sqrt{9}$$

Simplify.

$$t-5=3 t-5=-3$$

Add 5 to both sides of the equation.

$$t=8 t=2$$