Solve 15=10t-5/4t^2

$$15 = 10 t - \frac { 5 } { 4 } t ^ { 2 }$$

Solve for t

$$t=2$$

$$t=6$$

Steps Using the Quadratic Formula

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

$$10t-\frac{5}{4}t^{2}=15$$

Subtract 15 from both sides.

$$10t-\frac{5}{4}t^{2}-15=0$$

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.

$$-\frac{5}{4}t^{2}+10t-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{5}{4} for a, 10 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-\frac{5}{4}\right)\left(-15\right)}}{2\left(-\frac{5}{4}\right)}$$

Square 10.

$$t=\frac{-10±\sqrt{100-4\left(-\frac{5}{4}\right)\left(-15\right)}}{2\left(-\frac{5}{4}\right)}$$

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

$$t=\frac{-10±\sqrt{100+5\left(-15\right)}}{2\left(-\frac{5}{4}\right)}$$

Multiply 5 times -15.

$$t=\frac{-10±\sqrt{100-75}}{2\left(-\frac{5}{4}\right)}$$

Add 100 to -75.

$$t=\frac{-10±\sqrt{25}}{2\left(-\frac{5}{4}\right)}$$

Take the square root of 25.

$$t=\frac{-10±5}{2\left(-\frac{5}{4}\right)}$$

Multiply 2 times -\frac{5}{4}.

$$t=\frac{-10±5}{-\frac{5}{2}}$$

Now solve the equation t=\frac{-10±5}{-\frac{5}{2}} when ± is plus. Add -10 to 5.

$$t=-\frac{5}{-\frac{5}{2}}$$

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

$$t=2$$

Now solve the equation t=\frac{-10±5}{-\frac{5}{2}} when ± is minus. Subtract 5 from -10.

$$t=-\frac{15}{-\frac{5}{2}}$$

Divide -15 by -\frac{5}{2} by multiplying -15 by the reciprocal of -\frac{5}{2}.

$$t=6$$

The equation is now solved.

$$t=2 t=6$$

Steps for Completing the Square

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

$$10t-\frac{5}{4}t^{2}=15$$

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.

$$-\frac{5}{4}t^{2}+10t=15$$

Divide both sides of the equation by -\frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{5}{4}t^{2}+10t}{-\frac{5}{4}}=\frac{15}{-\frac{5}{4}}$$

Dividing by -\frac{5}{4} undoes the multiplication by -\frac{5}{4}.

$$t^{2}+\frac{10}{-\frac{5}{4}}t=\frac{15}{-\frac{5}{4}}$$

Divide 10 by -\frac{5}{4} by multiplying 10 by the reciprocal of -\frac{5}{4}.

$$t^{2}-8t=\frac{15}{-\frac{5}{4}}$$

Divide 15 by -\frac{5}{4} by multiplying 15 by the reciprocal of -\frac{5}{4}.

$$t^{2}-8t=-12$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-8t+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Square -4.

$$t^{2}-8t+16=-12+16$$

Add -12 to 16.

$$t^{2}-8t+16=4$$

Factor t^{2}-8t+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$t-4=2 t-4=-2$$

Add 4 to both sides of the equation.

$$t=6 t=2$$