Solve 0=-t^2+5t-5

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

Solve for t

$$t = \frac{\sqrt{5} + 5}{2} \approx 3.618033989$$

$$t = \frac{5 - \sqrt{5}}{2} \approx 1.381966011$$

Steps Using the Quadratic Formula

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Square 5.

$$t=\frac{-5±\sqrt{25-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-5±\sqrt{25+4\left(-5\right)}}{2\left(-1\right)}$$

Multiply 4 times -5.

$$t=\frac{-5±\sqrt{25-20}}{2\left(-1\right)}$$

Add 25 to -20.

$$t=\frac{-5±\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{-5±\sqrt{5}}{-2}$$

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

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

Divide -5+\sqrt{5} by -2.

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

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

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

Divide -5-\sqrt{5} by -2.

$$t=\frac{\sqrt{5}+5}{2}$$

The equation is now solved.

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

Steps for Completing the Square

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

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

Add 5 to both sides. Anything plus zero gives itself.

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

Divide both sides by -1.

$$\frac{-t^{2}+5t}{-1}=\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{5}{-1}t=\frac{5}{-1}$$

Divide 5 by -1.

$$t^{2}-5t=\frac{5}{-1}$$

Divide 5 by -1.

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

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-5t+\frac{25}{4}=-5+\frac{25}{4}$$

Add -5 to \frac{25}{4}.

$$t^{2}-5t+\frac{25}{4}=\frac{5}{4}$$

Factor t^{2}-5t+\frac{25}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{5}{2} to both sides of the equation.

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