Solve 84=5t+3/2t^2

$$84 = 5 t + \frac { 3 } { 2 } t ^ { 2 }$$

Solve for t

$$t = -\frac{28}{3} = -9\frac{1}{3} \approx -9.333333333$$

$$t=6$$

Steps Using the Quadratic Formula

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

$$5t+\frac{3}{2}t^{2}=84$$

Subtract 84 from both sides.

$$5t+\frac{3}{2}t^{2}-84=0$$

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.

$$\frac{3}{2}t^{2}+5t-84=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{2} for a, 5 for b, and -84 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-5±\sqrt{5^{2}-4\times \frac{3}{2}\left(-84\right)}}{2\times \frac{3}{2}}$$

Square 5.

$$t=\frac{-5±\sqrt{25-4\times \frac{3}{2}\left(-84\right)}}{2\times \frac{3}{2}}$$

Multiply -4 times \frac{3}{2}.

$$t=\frac{-5±\sqrt{25-6\left(-84\right)}}{2\times \frac{3}{2}}$$

Multiply -6 times -84.

$$t=\frac{-5±\sqrt{25+504}}{2\times \frac{3}{2}}$$

Add 25 to 504.

$$t=\frac{-5±\sqrt{529}}{2\times \frac{3}{2}}$$

Take the square root of 529.

$$t=\frac{-5±23}{2\times \frac{3}{2}}$$

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

$$t=\frac{-5±23}{3}$$

Now solve the equation t=\frac{-5±23}{3} when ± is plus. Add -5 to 23.

$$t=\frac{18}{3}$$

Divide 18 by 3.

$$t=6$$

Now solve the equation t=\frac{-5±23}{3} when ± is minus. Subtract 23 from -5.

$$t=-\frac{28}{3}$$

The equation is now solved.

$$t=6 t=-\frac{28}{3}$$

Steps for Completing the Square

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

$$5t+\frac{3}{2}t^{2}=84$$

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.

$$\frac{3}{2}t^{2}+5t=84$$

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

$$\frac{\frac{3}{2}t^{2}+5t}{\frac{3}{2}}=\frac{84}{\frac{3}{2}}$$

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

$$t^{2}+\frac{5}{\frac{3}{2}}t=\frac{84}{\frac{3}{2}}$$

Divide 5 by \frac{3}{2} by multiplying 5 by the reciprocal of \frac{3}{2}.

$$t^{2}+\frac{10}{3}t=\frac{84}{\frac{3}{2}}$$

Divide 84 by \frac{3}{2} by multiplying 84 by the reciprocal of \frac{3}{2}.

$$t^{2}+\frac{10}{3}t=56$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{10}{3}t+\left(\frac{5}{3}\right)^{2}=56+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{10}{3}t+\frac{25}{9}=56+\frac{25}{9}$$

Add 56 to \frac{25}{9}.

$$t^{2}+\frac{10}{3}t+\frac{25}{9}=\frac{529}{9}$$

Factor t^{2}+\frac{10}{3}t+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{5}{3}\right)^{2}=\frac{529}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{5}{3}\right)^{2}}=\sqrt{\frac{529}{9}}$$

Simplify.

$$t+\frac{5}{3}=\frac{23}{3} t+\frac{5}{3}=-\frac{23}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$t=6 t=-\frac{28}{3}$$