Solve 6x^2+4x-32x+7/2=9

$$\frac { 6 x ^ { 2 } + 4 x - 32 x + 7 } { 2 } = 9$$

Solve for x

$$x=\frac{\sqrt{262}}{6}+\frac{7}{3}\approx 5.031069009$$

$$x=-\frac{\sqrt{262}}{6}+\frac{7}{3}\approx -0.364402343$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$6x^{2}+4x-32x+7=9\times 2$$

Combine 4x and -32x to get -28x.

$$6x^{2}-28x+7=9\times 2$$

Multiply 9 and 2 to get 18.

$$6x^{2}-28x+7=18$$

Subtract 18 from both sides.

$$6x^{2}-28x+7-18=0$$

Subtract 18 from 7 to get -11.

$$6x^{2}-28x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -28 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 6\left(-11\right)}}{2\times 6}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 6\left(-11\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-28\right)±\sqrt{784-24\left(-11\right)}}{2\times 6}$$

Multiply -24 times -11.

$$x=\frac{-\left(-28\right)±\sqrt{784+264}}{2\times 6}$$

Add 784 to 264.

$$x=\frac{-\left(-28\right)±\sqrt{1048}}{2\times 6}$$

Take the square root of 1048.

$$x=\frac{-\left(-28\right)±2\sqrt{262}}{2\times 6}$$

The opposite of -28 is 28.

$$x=\frac{28±2\sqrt{262}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{28±2\sqrt{262}}{12}$$

Now solve the equation x=\frac{28±2\sqrt{262}}{12} when ± is plus. Add 28 to 2\sqrt{262}.

$$x=\frac{2\sqrt{262}+28}{12}$$

Divide 28+2\sqrt{262} by 12.

$$x=\frac{\sqrt{262}}{6}+\frac{7}{3}$$

Now solve the equation x=\frac{28±2\sqrt{262}}{12} when ± is minus. Subtract 2\sqrt{262} from 28.

$$x=\frac{28-2\sqrt{262}}{12}$$

Divide 28-2\sqrt{262} by 12.

$$x=-\frac{\sqrt{262}}{6}+\frac{7}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{262}}{6}+\frac{7}{3} x=-\frac{\sqrt{262}}{6}+\frac{7}{3}$$

Steps for Completing the Square

Multiply both sides by 2.

$$6x^{2}+4x-32x+7=9\times 2$$

Combine 4x and -32x to get -28x.

$$6x^{2}-28x+7=9\times 2$$

Multiply 9 and 2 to get 18.

$$6x^{2}-28x+7=18$$

Subtract 7 from both sides.

$$6x^{2}-28x=18-7$$

Subtract 7 from 18 to get 11.

$$6x^{2}-28x=11$$

Divide both sides by 6.

$$\frac{6x^{2}-28x}{6}=\frac{11}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{28}{6}\right)x=\frac{11}{6}$$

Reduce the fraction \frac{-28}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{14}{3}x=\frac{11}{6}$$

Divide -\frac{14}{3}, the coefficient of the x term, by 2 to get -\frac{7}{3}. Then add the square of -\frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{14}{3}x+\left(-\frac{7}{3}\right)^{2}=\frac{11}{6}+\left(-\frac{7}{3}\right)^{2}$$

Square -\frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{11}{6}+\frac{49}{9}$$

Add \frac{11}{6} to \frac{49}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{131}{18}$$

Factor x^{2}-\frac{14}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{3}\right)^{2}=\frac{131}{18}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{3}\right)^{2}}=\sqrt{\frac{131}{18}}$$

Simplify.

$$x-\frac{7}{3}=\frac{\sqrt{262}}{6} x-\frac{7}{3}=-\frac{\sqrt{262}}{6}$$

Add \frac{7}{3} to both sides of the equation.

$$x=\frac{\sqrt{262}}{6}+\frac{7}{3} x=-\frac{\sqrt{262}}{6}+\frac{7}{3}$$