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5q^{2}-7q=0
Subtract 7q from both sides.
q\left(5q-7\right)=0
Factor out q.
q=0 q=\frac{7}{5}
To find equation solutions, solve q=0 and 5q-7=0.
5q^{2}-7q=0
Subtract 7q from both sides.
q=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -7 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
q=\frac{-\left(-7\right)±7}{2\times 5}
Take the square root of \left(-7\right)^{2}.
q=\frac{7±7}{2\times 5}
The opposite of -7 is 7.
q=\frac{7±7}{10}
Multiply 2 times 5.
q=\frac{14}{10}
Now solve the equation q=\frac{7±7}{10} when ± is plus. Add 7 to 7.
q=\frac{7}{5}
Reduce the fraction \frac{14}{10} to lowest terms by extracting and canceling out 2.
q=\frac{0}{10}
Now solve the equation q=\frac{7±7}{10} when ± is minus. Subtract 7 from 7.
q=0
Divide 0 by 10.
q=\frac{7}{5} q=0
The equation is now solved.
5q^{2}-7q=0
Subtract 7q from both sides.
\frac{5q^{2}-7q}{5}=\frac{0}{5}
Divide both sides by 5.
q^{2}-\frac{7}{5}q=\frac{0}{5}
Dividing by 5 undoes the multiplication by 5.
q^{2}-\frac{7}{5}q=0
Divide 0 by 5.
q^{2}-\frac{7}{5}q+\left(-\frac{7}{10}\right)^{2}=\left(-\frac{7}{10}\right)^{2}
Divide -\frac{7}{5}, the coefficient of the x term, by 2 to get -\frac{7}{10}. Then add the square of -\frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
q^{2}-\frac{7}{5}q+\frac{49}{100}=\frac{49}{100}
Square -\frac{7}{10} by squaring both the numerator and the denominator of the fraction.
\left(q-\frac{7}{10}\right)^{2}=\frac{49}{100}
Factor q^{2}-\frac{7}{5}q+\frac{49}{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}.
\sqrt{\left(q-\frac{7}{10}\right)^{2}}=\sqrt{\frac{49}{100}}
Take the square root of both sides of the equation.
q-\frac{7}{10}=\frac{7}{10} q-\frac{7}{10}=-\frac{7}{10}
Simplify.
q=\frac{7}{5} q=0
Add \frac{7}{10} to both sides of the equation.