Skip to main content
Solve for s
Tick mark Image

Similar Problems from Web Search

Share

s\left(s+6\right)=0
Factor out s.
s=0 s=-6
To find equation solutions, solve s=0 and s+6=0.
s^{2}+6s=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.
s=\frac{-6±\sqrt{6^{2}}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
s=\frac{-6±6}{2}
Take the square root of 6^{2}.
s=\frac{0}{2}
Now solve the equation s=\frac{-6±6}{2} when ± is plus. Add -6 to 6.
s=0
Divide 0 by 2.
s=-\frac{12}{2}
Now solve the equation s=\frac{-6±6}{2} when ± is minus. Subtract 6 from -6.
s=-6
Divide -12 by 2.
s=0 s=-6
The equation is now solved.
s^{2}+6s=0
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.
s^{2}+6s+3^{2}=3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
s^{2}+6s+9=9
Square 3.
\left(s+3\right)^{2}=9
Factor s^{2}+6s+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(s+3\right)^{2}}=\sqrt{9}
Take the square root of both sides of the equation.
s+3=3 s+3=-3
Simplify.
s=0 s=-6
Subtract 3 from both sides of the equation.