Solve 6x^2=8x+7

$$6 x ^ { 2 } = 8 x + 7$$

Solve for x

$$x=\frac{\sqrt{58}}{6}+\frac{2}{3}\approx 1.935962184$$

$$x=-\frac{\sqrt{58}}{6}+\frac{2}{3}\approx -0.602628851$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$6x^{2}-8x=7$$

Subtract 7 from both sides.

$$6x^{2}-8x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -8 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 6\left(-7\right)}}{2\times 6}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 6\left(-7\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-8\right)±\sqrt{64-24\left(-7\right)}}{2\times 6}$$

Multiply -24 times -7.

$$x=\frac{-\left(-8\right)±\sqrt{64+168}}{2\times 6}$$

Add 64 to 168.

$$x=\frac{-\left(-8\right)±\sqrt{232}}{2\times 6}$$

Take the square root of 232.

$$x=\frac{-\left(-8\right)±2\sqrt{58}}{2\times 6}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{58}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{8±2\sqrt{58}}{12}$$

Now solve the equation x=\frac{8±2\sqrt{58}}{12} when ± is plus. Add 8 to 2\sqrt{58}.

$$x=\frac{2\sqrt{58}+8}{12}$$

Divide 8+2\sqrt{58} by 12.

$$x=\frac{\sqrt{58}}{6}+\frac{2}{3}$$

Now solve the equation x=\frac{8±2\sqrt{58}}{12} when ± is minus. Subtract 2\sqrt{58} from 8.

$$x=\frac{8-2\sqrt{58}}{12}$$

Divide 8-2\sqrt{58} by 12.

$$x=-\frac{\sqrt{58}}{6}+\frac{2}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{58}}{6}+\frac{2}{3} x=-\frac{\sqrt{58}}{6}+\frac{2}{3}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$6x^{2}-8x=7$$

Divide both sides by 6.

$$\frac{6x^{2}-8x}{6}=\frac{7}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{8}{6}\right)x=\frac{7}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{4}{3}x=\frac{7}{6}$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{7}{6}+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{7}{6}+\frac{4}{9}$$

Add \frac{7}{6} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{29}{18}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{29}{18}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{29}{18}}$$

Simplify.

$$x-\frac{2}{3}=\frac{\sqrt{58}}{6} x-\frac{2}{3}=-\frac{\sqrt{58}}{6}$$

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{\sqrt{58}}{6}+\frac{2}{3} x=-\frac{\sqrt{58}}{6}+\frac{2}{3}$$