Solve b^2+15b+5b=0

$$b ^ { 2 } + 15 b + 5 b = 0$$

Solve for b

$$b=-20$$

$$b=0$$

Steps Using Factoring

Factor out b.

$$b\left(b+15+5\right)=0$$

To find equation solutions, solve b=0 and b+20=0.

$$b=0 b=-20$$

Steps Using the Quadratic Formula

Combine 15b and 5b to get 20b.

$$b^{2}+20b=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-20±\sqrt{20^{2}}}{2}$$

Take the square root of 20^{2}.

$$b=\frac{-20±20}{2}$$

Now solve the equation b=\frac{-20±20}{2} when ± is plus. Add -20 to 20.

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

Divide 0 by 2.

$$b=0$$

Now solve the equation b=\frac{-20±20}{2} when ± is minus. Subtract 20 from -20.

$$b=-\frac{40}{2}$$

Divide -40 by 2.

$$b=-20$$

The equation is now solved.

$$b=0 b=-20$$

Steps for Completing the Square

Combine 15b and 5b to get 20b.

$$b^{2}+20b=0$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}+20b+10^{2}=10^{2}$$

Square 10.

$$b^{2}+20b+100=100$$

Factor b^{2}+20b+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b+10\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+10\right)^{2}}=\sqrt{100}$$

Simplify.

$$b+10=10 b+10=-10$$

Subtract 10 from both sides of the equation.

$$b=0 b=-20$$