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5q^{2}+15q+5=-6
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.
5q^{2}+15q+5-\left(-6\right)=-6-\left(-6\right)
Add 6 to both sides of the equation.
5q^{2}+15q+5-\left(-6\right)=0
Subtracting -6 from itself leaves 0.
5q^{2}+15q+11=0
Subtract -6 from 5.
q=\frac{-15±\sqrt{15^{2}-4\times 5\times 11}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 15 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
q=\frac{-15±\sqrt{225-4\times 5\times 11}}{2\times 5}
Square 15.
q=\frac{-15±\sqrt{225-20\times 11}}{2\times 5}
Multiply -4 times 5.
q=\frac{-15±\sqrt{225-220}}{2\times 5}
Multiply -20 times 11.
q=\frac{-15±\sqrt{5}}{2\times 5}
Add 225 to -220.
q=\frac{-15±\sqrt{5}}{10}
Multiply 2 times 5.
q=\frac{\sqrt{5}-15}{10}
Now solve the equation q=\frac{-15±\sqrt{5}}{10} when ± is plus. Add -15 to \sqrt{5}.
q=\frac{\sqrt{5}}{10}-\frac{3}{2}
Divide -15+\sqrt{5} by 10.
q=\frac{-\sqrt{5}-15}{10}
Now solve the equation q=\frac{-15±\sqrt{5}}{10} when ± is minus. Subtract \sqrt{5} from -15.
q=-\frac{\sqrt{5}}{10}-\frac{3}{2}
Divide -15-\sqrt{5} by 10.
q=\frac{\sqrt{5}}{10}-\frac{3}{2} q=-\frac{\sqrt{5}}{10}-\frac{3}{2}
The equation is now solved.
5q^{2}+15q+5=-6
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.
5q^{2}+15q+5-5=-6-5
Subtract 5 from both sides of the equation.
5q^{2}+15q=-6-5
Subtracting 5 from itself leaves 0.
5q^{2}+15q=-11
Subtract 5 from -6.
\frac{5q^{2}+15q}{5}=-\frac{11}{5}
Divide both sides by 5.
q^{2}+\frac{15}{5}q=-\frac{11}{5}
Dividing by 5 undoes the multiplication by 5.
q^{2}+3q=-\frac{11}{5}
Divide 15 by 5.
q^{2}+3q+\left(\frac{3}{2}\right)^{2}=-\frac{11}{5}+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
q^{2}+3q+\frac{9}{4}=-\frac{11}{5}+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
q^{2}+3q+\frac{9}{4}=\frac{1}{20}
Add -\frac{11}{5} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(q+\frac{3}{2}\right)^{2}=\frac{1}{20}
Factor q^{2}+3q+\frac{9}{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(q+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{20}}
Take the square root of both sides of the equation.
q+\frac{3}{2}=\frac{\sqrt{5}}{10} q+\frac{3}{2}=-\frac{\sqrt{5}}{10}
Simplify.
q=\frac{\sqrt{5}}{10}-\frac{3}{2} q=-\frac{\sqrt{5}}{10}-\frac{3}{2}
Subtract \frac{3}{2} from both sides of the equation.