Solve 6x^2=14x-5

$$6 { x }^{ 2 } =14x-5$$

Solve for x

$$x = \frac{\sqrt{19} + 7}{6} \approx 1.893149824$$

$$x=\frac{7-\sqrt{19}}{6}\approx 0.440183509$$

Steps Using the Quadratic Formula

Subtract 14x from both sides.

$$6x^{2}-14x=-5$$

Add 5 to both sides.

$$6x^{2}-14x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -14 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 6\times 5}}{2\times 6}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 6\times 5}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-14\right)±\sqrt{196-24\times 5}}{2\times 6}$$

Multiply -24 times 5.

$$x=\frac{-\left(-14\right)±\sqrt{196-120}}{2\times 6}$$

Add 196 to -120.

$$x=\frac{-\left(-14\right)±\sqrt{76}}{2\times 6}$$

Take the square root of 76.

$$x=\frac{-\left(-14\right)±2\sqrt{19}}{2\times 6}$$

The opposite of -14 is 14.

$$x=\frac{14±2\sqrt{19}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{14±2\sqrt{19}}{12}$$

Now solve the equation x=\frac{14±2\sqrt{19}}{12} when ± is plus. Add 14 to 2\sqrt{19}.

$$x=\frac{2\sqrt{19}+14}{12}$$

Divide 14+2\sqrt{19} by 12.

$$x=\frac{\sqrt{19}+7}{6}$$

Now solve the equation x=\frac{14±2\sqrt{19}}{12} when ± is minus. Subtract 2\sqrt{19} from 14.

$$x=\frac{14-2\sqrt{19}}{12}$$

Divide 14-2\sqrt{19} by 12.

$$x=\frac{7-\sqrt{19}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{19}+7}{6} x=\frac{7-\sqrt{19}}{6}$$

Steps for Completing the Square

Subtract 14x from both sides.

$$6x^{2}-14x=-5$$

Divide both sides by 6.

$$\frac{6x^{2}-14x}{6}=-\frac{5}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{14}{6}\right)x=-\frac{5}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{7}{3}x=-\frac{5}{6}$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=-\frac{5}{6}+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{5}{6}+\frac{49}{36}$$

Add -\frac{5}{6} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{19}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{19}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{19}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{\sqrt{19}}{6} x-\frac{7}{6}=-\frac{\sqrt{19}}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=\frac{\sqrt{19}+7}{6} x=\frac{7-\sqrt{19}}{6}$$