Solve 6x^2+7x+9=11x^2

$$6 x ^ { 2 } + 7 x + 9 = 11 x ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{229} + 7}{10} \approx 2.213274595$$

$$x=\frac{7-\sqrt{229}}{10}\approx -0.813274595$$

Steps Using the Quadratic Formula

Subtract 11x^{2} from both sides.

$$6x^{2}+7x+9-11x^{2}=0$$

Combine 6x^{2} and -11x^{2} to get -5x^{2}.

$$-5x^{2}+7x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 7 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-5\right)\times 9}}{2\left(-5\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-5\right)\times 9}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-7±\sqrt{49+20\times 9}}{2\left(-5\right)}$$

Multiply 20 times 9.

$$x=\frac{-7±\sqrt{49+180}}{2\left(-5\right)}$$

Add 49 to 180.

$$x=\frac{-7±\sqrt{229}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-7±\sqrt{229}}{-10}$$

Now solve the equation x=\frac{-7±\sqrt{229}}{-10} when ± is plus. Add -7 to \sqrt{229}.

$$x=\frac{\sqrt{229}-7}{-10}$$

Divide -7+\sqrt{229} by -10.

$$x=\frac{7-\sqrt{229}}{10}$$

Now solve the equation x=\frac{-7±\sqrt{229}}{-10} when ± is minus. Subtract \sqrt{229} from -7.

$$x=\frac{-\sqrt{229}-7}{-10}$$

Divide -7-\sqrt{229} by -10.

$$x=\frac{\sqrt{229}+7}{10}$$

The equation is now solved.

$$x=\frac{7-\sqrt{229}}{10} x=\frac{\sqrt{229}+7}{10}$$

Steps for Completing the Square

Subtract 11x^{2} from both sides.

$$6x^{2}+7x+9-11x^{2}=0$$

Combine 6x^{2} and -11x^{2} to get -5x^{2}.

$$-5x^{2}+7x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}+7x=-9$$

Divide both sides by -5.

$$\frac{-5x^{2}+7x}{-5}=-\frac{9}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{7}{-5}x=-\frac{9}{-5}$$

Divide 7 by -5.

$$x^{2}-\frac{7}{5}x=-\frac{9}{-5}$$

Divide -9 by -5.

$$x^{2}-\frac{7}{5}x=\frac{9}{5}$$

Divide -\frac{7}{5}, the coefficient of the x term, by 2 to get -\frac{7}{10}. Then add the square of -\frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{5}x+\left(-\frac{7}{10}\right)^{2}=\frac{9}{5}+\left(-\frac{7}{10}\right)^{2}$$

Square -\frac{7}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{9}{5}+\frac{49}{100}$$

Add \frac{9}{5} to \frac{49}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{229}{100}$$

Factor x^{2}-\frac{7}{5}x+\frac{49}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{10}\right)^{2}=\frac{229}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{10}\right)^{2}}=\sqrt{\frac{229}{100}}$$

Simplify.

$$x-\frac{7}{10}=\frac{\sqrt{229}}{10} x-\frac{7}{10}=-\frac{\sqrt{229}}{10}$$

Add \frac{7}{10} to both sides of the equation.

$$x=\frac{\sqrt{229}+7}{10} x=\frac{7-\sqrt{229}}{10}$$