Solve b^2=5b

$$b ^ { 2 } = 5 b$$

Solve for b

$$b=5$$

$$b=0$$

Steps Using Factoring

Subtract 5b from both sides.

$$b^{2}-5b=0$$

Factor out b.

$$b\left(b-5\right)=0$$

To find equation solutions, solve b=0 and b-5=0.

$$b=0 b=5$$

Steps Using the Quadratic Formula

Subtract 5b from both sides.

$$b^{2}-5b=0$$

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

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

Take the square root of \left(-5\right)^{2}.

$$b=\frac{-\left(-5\right)±5}{2}$$

The opposite of -5 is 5.

$$b=\frac{5±5}{2}$$

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

$$b=\frac{10}{2}$$

Divide 10 by 2.

$$b=5$$

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

$$b=\frac{0}{2}$$

Divide 0 by 2.

$$b=0$$

The equation is now solved.

$$b=5 b=0$$

Steps for Completing the Square

Subtract 5b from both sides.

$$b^{2}-5b=0$$

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.

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

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

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

Factor b^{2}-5b+\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}.

$$\left(b-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

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

$$b=5 b=0$$