Solve 10=5t-5t^2

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

Solve for t

$$t=\frac{1+\sqrt{7}i}{2}\approx 0.5+1.322875656i$$

$$t=\frac{-\sqrt{7}i+1}{2}\approx 0.5-1.322875656i$$

Steps Using the Quadratic Formula

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

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

Subtract 10 from both sides.

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

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.

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

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

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

Square 5.

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

Multiply -4 times -5.

$$t=\frac{-5±\sqrt{25+20\left(-10\right)}}{2\left(-5\right)}$$

Multiply 20 times -10.

$$t=\frac{-5±\sqrt{25-200}}{2\left(-5\right)}$$

Add 25 to -200.

$$t=\frac{-5±\sqrt{-175}}{2\left(-5\right)}$$

Take the square root of -175.

$$t=\frac{-5±5\sqrt{7}i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-5±5\sqrt{7}i}{-10}$$

Now solve the equation t=\frac{-5±5\sqrt{7}i}{-10} when ± is plus. Add -5 to 5i\sqrt{7}.

$$t=\frac{-5+5\sqrt{7}i}{-10}$$

Divide -5+5i\sqrt{7} by -10.

$$t=\frac{-\sqrt{7}i+1}{2}$$

Now solve the equation t=\frac{-5±5\sqrt{7}i}{-10} when ± is minus. Subtract 5i\sqrt{7} from -5.

$$t=\frac{-5\sqrt{7}i-5}{-10}$$

Divide -5-5i\sqrt{7} by -10.

$$t=\frac{1+\sqrt{7}i}{2}$$

The equation is now solved.

$$t=\frac{-\sqrt{7}i+1}{2} t=\frac{1+\sqrt{7}i}{2}$$

Steps for Completing the Square

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

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

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.

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

Divide both sides by -5.

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

Dividing by -5 undoes the multiplication by -5.

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

Divide 5 by -5.

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

Divide 10 by -5.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-t+\frac{1}{4}=-2+\frac{1}{4}$$

Add -2 to \frac{1}{4}.

$$t^{2}-t+\frac{1}{4}=-\frac{7}{4}$$

Factor t^{2}-t+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{1}{2}\right)^{2}=-\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Simplify.

$$t-\frac{1}{2}=\frac{\sqrt{7}i}{2} t-\frac{1}{2}=-\frac{\sqrt{7}i}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$t=\frac{1+\sqrt{7}i}{2} t=\frac{-\sqrt{7}i+1}{2}$$