Solve 5r^2+2=12r

$$5 r ^ { 2 } + 2 = 12 r$$

Solve for r

$$r = \frac{\sqrt{26} + 6}{5} \approx 2.219803903$$

$$r=\frac{6-\sqrt{26}}{5}\approx 0.180196097$$

Steps Using the Quadratic Formula

Subtract 12r from both sides.

$$5r^{2}+2-12r=0$$

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.

$$5r^{2}-12r+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -12 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\times 2}}{2\times 5}$$

Square -12.

$$r=\frac{-\left(-12\right)±\sqrt{144-4\times 5\times 2}}{2\times 5}$$

Multiply -4 times 5.

$$r=\frac{-\left(-12\right)±\sqrt{144-20\times 2}}{2\times 5}$$

Multiply -20 times 2.

$$r=\frac{-\left(-12\right)±\sqrt{144-40}}{2\times 5}$$

Add 144 to -40.

$$r=\frac{-\left(-12\right)±\sqrt{104}}{2\times 5}$$

Take the square root of 104.

$$r=\frac{-\left(-12\right)±2\sqrt{26}}{2\times 5}$$

The opposite of -12 is 12.

$$r=\frac{12±2\sqrt{26}}{2\times 5}$$

Multiply 2 times 5.

$$r=\frac{12±2\sqrt{26}}{10}$$

Now solve the equation r=\frac{12±2\sqrt{26}}{10} when ± is plus. Add 12 to 2\sqrt{26}.

$$r=\frac{2\sqrt{26}+12}{10}$$

Divide 12+2\sqrt{26} by 10.

$$r=\frac{\sqrt{26}+6}{5}$$

Now solve the equation r=\frac{12±2\sqrt{26}}{10} when ± is minus. Subtract 2\sqrt{26} from 12.

$$r=\frac{12-2\sqrt{26}}{10}$$

Divide 12-2\sqrt{26} by 10.

$$r=\frac{6-\sqrt{26}}{5}$$

The equation is now solved.

$$r=\frac{\sqrt{26}+6}{5} r=\frac{6-\sqrt{26}}{5}$$

Steps for Completing the Square

Subtract 12r from both sides.

$$5r^{2}+2-12r=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$5r^{2}-12r=-2$$

Divide both sides by 5.

$$\frac{5r^{2}-12r}{5}=-\frac{2}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$r^{2}-\frac{12}{5}r=-\frac{2}{5}$$

Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-\frac{12}{5}r+\left(-\frac{6}{5}\right)^{2}=-\frac{2}{5}+\left(-\frac{6}{5}\right)^{2}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-\frac{12}{5}r+\frac{36}{25}=-\frac{2}{5}+\frac{36}{25}$$

Add -\frac{2}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r^{2}-\frac{12}{5}r+\frac{36}{25}=\frac{26}{25}$$

Factor r^{2}-\frac{12}{5}r+\frac{36}{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}.

$$\left(r-\frac{6}{5}\right)^{2}=\frac{26}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{6}{5}\right)^{2}}=\sqrt{\frac{26}{25}}$$

Simplify.

$$r-\frac{6}{5}=\frac{\sqrt{26}}{5} r-\frac{6}{5}=-\frac{\sqrt{26}}{5}$$

Add \frac{6}{5} to both sides of the equation.

$$r=\frac{\sqrt{26}+6}{5} r=\frac{6-\sqrt{26}}{5}$$