Solve 5q^2=14q

$$5 q ^ { 2 } = 14 q$$

Solve for q

$$q = \frac{14}{5} = 2\frac{4}{5} = 2.8$$

$$q=0$$

Steps Using Factoring

Subtract 14q from both sides.

$$5q^{2}-14q=0$$

Factor out q.

$$q\left(5q-14\right)=0$$

To find equation solutions, solve q=0 and 5q-14=0.

$$q=0 q=\frac{14}{5}$$

Steps Using the Quadratic Formula

Subtract 14q from both sides.

$$5q^{2}-14q=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -14 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}}}{2\times 5}$$

Take the square root of \left(-14\right)^{2}.

$$q=\frac{-\left(-14\right)±14}{2\times 5}$$

The opposite of -14 is 14.

$$q=\frac{14±14}{2\times 5}$$

Multiply 2 times 5.

$$q=\frac{14±14}{10}$$

Now solve the equation q=\frac{14±14}{10} when ± is plus. Add 14 to 14.

$$q=\frac{28}{10}$$

Reduce the fraction \frac{28}{10} to lowest terms by extracting and canceling out 2.

$$q=\frac{14}{5}$$

Now solve the equation q=\frac{14±14}{10} when ± is minus. Subtract 14 from 14.

$$q=\frac{0}{10}$$

Divide 0 by 10.

$$q=0$$

The equation is now solved.

$$q=\frac{14}{5} q=0$$

Steps for Completing the Square

Subtract 14q from both sides.

$$5q^{2}-14q=0$$

Divide both sides by 5.

$$\frac{5q^{2}-14q}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$q^{2}-\frac{14}{5}q=\frac{0}{5}$$

Divide 0 by 5.

$$q^{2}-\frac{14}{5}q=0$$

Divide -\frac{14}{5}, the coefficient of the x term, by 2 to get -\frac{7}{5}. Then add the square of -\frac{7}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}-\frac{14}{5}q+\left(-\frac{7}{5}\right)^{2}=\left(-\frac{7}{5}\right)^{2}$$

Square -\frac{7}{5} by squaring both the numerator and the denominator of the fraction.

$$q^{2}-\frac{14}{5}q+\frac{49}{25}=\frac{49}{25}$$

Factor q^{2}-\frac{14}{5}q+\frac{49}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q-\frac{7}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q-\frac{7}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

$$q-\frac{7}{5}=\frac{7}{5} q-\frac{7}{5}=-\frac{7}{5}$$

Add \frac{7}{5} to both sides of the equation.

$$q=\frac{14}{5} q=0$$