Solve 7q^2=q+2

$$7 q ^ { 2 } = q + 2$$

Solve for q

$$q=\frac{\sqrt{57}+1}{14}\approx 0.61070246$$

$$q=\frac{1-\sqrt{57}}{14}\approx -0.467845317$$

Steps Using the Quadratic Formula

Subtract q from both sides.

$$7q^{2}-q=2$$

Subtract 2 from both sides.

$$7q^{2}-q-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -1 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-1\right)±\sqrt{1-4\times 7\left(-2\right)}}{2\times 7}$$

Multiply -4 times 7.

$$q=\frac{-\left(-1\right)±\sqrt{1-28\left(-2\right)}}{2\times 7}$$

Multiply -28 times -2.

$$q=\frac{-\left(-1\right)±\sqrt{1+56}}{2\times 7}$$

Add 1 to 56.

$$q=\frac{-\left(-1\right)±\sqrt{57}}{2\times 7}$$

The opposite of -1 is 1.

$$q=\frac{1±\sqrt{57}}{2\times 7}$$

Multiply 2 times 7.

$$q=\frac{1±\sqrt{57}}{14}$$

Now solve the equation q=\frac{1±\sqrt{57}}{14} when ± is plus. Add 1 to \sqrt{57}.

$$q=\frac{\sqrt{57}+1}{14}$$

Now solve the equation q=\frac{1±\sqrt{57}}{14} when ± is minus. Subtract \sqrt{57} from 1.

$$q=\frac{1-\sqrt{57}}{14}$$

The equation is now solved.

$$q=\frac{\sqrt{57}+1}{14} q=\frac{1-\sqrt{57}}{14}$$

Steps for Completing the Square

Subtract q from both sides.

$$7q^{2}-q=2$$

Divide both sides by 7.

$$\frac{7q^{2}-q}{7}=\frac{2}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$q^{2}-\frac{1}{7}q=\frac{2}{7}$$

Divide -\frac{1}{7}, the coefficient of the x term, by 2 to get -\frac{1}{14}. Then add the square of -\frac{1}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}-\frac{1}{7}q+\left(-\frac{1}{14}\right)^{2}=\frac{2}{7}+\left(-\frac{1}{14}\right)^{2}$$

Square -\frac{1}{14} by squaring both the numerator and the denominator of the fraction.

$$q^{2}-\frac{1}{7}q+\frac{1}{196}=\frac{2}{7}+\frac{1}{196}$$

Add \frac{2}{7} to \frac{1}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$q^{2}-\frac{1}{7}q+\frac{1}{196}=\frac{57}{196}$$

Factor q^{2}-\frac{1}{7}q+\frac{1}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q-\frac{1}{14}\right)^{2}=\frac{57}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q-\frac{1}{14}\right)^{2}}=\sqrt{\frac{57}{196}}$$

Simplify.

$$q-\frac{1}{14}=\frac{\sqrt{57}}{14} q-\frac{1}{14}=-\frac{\sqrt{57}}{14}$$

Add \frac{1}{14} to both sides of the equation.

$$q=\frac{\sqrt{57}+1}{14} q=\frac{1-\sqrt{57}}{14}$$