Solve 65=-10t-5t^2

$$65 = -10t-5 { t }^{ 2 }$$

Solve for t

$$t=-1+2\sqrt{3}i\approx -1+3.464101615i$$

$$t=-2\sqrt{3}i-1\approx -1-3.464101615i$$

Steps Using the Quadratic Formula

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

$$-10t-5t^{2}=65$$

Subtract 65 from both sides.

$$-10t-5t^{2}-65=0$$

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.

$$-5t^{2}-10t-65=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -10 for b, and -65 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-5\right)\left(-65\right)}}{2\left(-5\right)}$$

Square -10.

$$t=\frac{-\left(-10\right)±\sqrt{100-4\left(-5\right)\left(-65\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-\left(-10\right)±\sqrt{100+20\left(-65\right)}}{2\left(-5\right)}$$

Multiply 20 times -65.

$$t=\frac{-\left(-10\right)±\sqrt{100-1300}}{2\left(-5\right)}$$

Add 100 to -1300.

$$t=\frac{-\left(-10\right)±\sqrt{-1200}}{2\left(-5\right)}$$

Take the square root of -1200.

$$t=\frac{-\left(-10\right)±20\sqrt{3}i}{2\left(-5\right)}$$

The opposite of -10 is 10.

$$t=\frac{10±20\sqrt{3}i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{10±20\sqrt{3}i}{-10}$$

Now solve the equation t=\frac{10±20\sqrt{3}i}{-10} when ± is plus. Add 10 to 20i\sqrt{3}.

$$t=\frac{10+20\sqrt{3}i}{-10}$$

Divide 10+20i\sqrt{3} by -10.

$$t=-2\sqrt{3}i-1$$

Now solve the equation t=\frac{10±20\sqrt{3}i}{-10} when ± is minus. Subtract 20i\sqrt{3} from 10.

$$t=\frac{-20\sqrt{3}i+10}{-10}$$

Divide 10-20i\sqrt{3} by -10.

$$t=-1+2\sqrt{3}i$$

The equation is now solved.

$$t=-2\sqrt{3}i-1 t=-1+2\sqrt{3}i$$

Steps for Completing the Square

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

$$-10t-5t^{2}=65$$

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.

$$-5t^{2}-10t=65$$

Divide both sides by -5.

$$\frac{-5t^{2}-10t}{-5}=\frac{65}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\left(-\frac{10}{-5}\right)t=\frac{65}{-5}$$

Divide -10 by -5.

$$t^{2}+2t=\frac{65}{-5}$$

Divide 65 by -5.

$$t^{2}+2t=-13$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+2t+1^{2}=-13+1^{2}$$

Square 1.

$$t^{2}+2t+1=-13+1$$

Add -13 to 1.

$$t^{2}+2t+1=-12$$

Factor t^{2}+2t+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+1\right)^{2}=-12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+1\right)^{2}}=\sqrt{-12}$$

Simplify.

$$t+1=2\sqrt{3}i t+1=-2\sqrt{3}i$$

Subtract 1 from both sides of the equation.

$$t=-1+2\sqrt{3}i t=-2\sqrt{3}i-1$$