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3u^{2}+15u=0
Add 15u to both sides.
u\left(3u+15\right)=0
Factor out u.
u=0 u=-5
To find equation solutions, solve u=0 and 3u+15=0.
3u^{2}+15u=0
Add 15u to both sides.
u=\frac{-15±\sqrt{15^{2}}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 15 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
u=\frac{-15±15}{2\times 3}
Take the square root of 15^{2}.
u=\frac{-15±15}{6}
Multiply 2 times 3.
u=\frac{0}{6}
Now solve the equation u=\frac{-15±15}{6} when ± is plus. Add -15 to 15.
u=0
Divide 0 by 6.
u=-\frac{30}{6}
Now solve the equation u=\frac{-15±15}{6} when ± is minus. Subtract 15 from -15.
u=-5
Divide -30 by 6.
u=0 u=-5
The equation is now solved.
3u^{2}+15u=0
Add 15u to both sides.
\frac{3u^{2}+15u}{3}=\frac{0}{3}
Divide both sides by 3.
u^{2}+\frac{15}{3}u=\frac{0}{3}
Dividing by 3 undoes the multiplication by 3.
u^{2}+5u=\frac{0}{3}
Divide 15 by 3.
u^{2}+5u=0
Divide 0 by 3.
u^{2}+5u+\left(\frac{5}{2}\right)^{2}=\left(\frac{5}{2}\right)^{2}
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.
u^{2}+5u+\frac{25}{4}=\frac{25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
\left(u+\frac{5}{2}\right)^{2}=\frac{25}{4}
Factor u^{2}+5u+\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}.
\sqrt{\left(u+\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}
Take the square root of both sides of the equation.
u+\frac{5}{2}=\frac{5}{2} u+\frac{5}{2}=-\frac{5}{2}
Simplify.
u=0 u=-5
Subtract \frac{5}{2} from both sides of the equation.