Solve 0=-9x^2+42x-11

$$0 = - 9 x ^ { 2 } + 42 x - 11$$

Solve for x

$$x = \frac{\sqrt{38} + 7}{3} \approx 4.388138001$$

$$x=\frac{7-\sqrt{38}}{3}\approx 0.278528666$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-9x^{2}+42x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 42 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-42±\sqrt{42^{2}-4\left(-9\right)\left(-11\right)}}{2\left(-9\right)}$$

Square 42.

$$x=\frac{-42±\sqrt{1764-4\left(-9\right)\left(-11\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-42±\sqrt{1764+36\left(-11\right)}}{2\left(-9\right)}$$

Multiply 36 times -11.

$$x=\frac{-42±\sqrt{1764-396}}{2\left(-9\right)}$$

Add 1764 to -396.

$$x=\frac{-42±\sqrt{1368}}{2\left(-9\right)}$$

Take the square root of 1368.

$$x=\frac{-42±6\sqrt{38}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-42±6\sqrt{38}}{-18}$$

Now solve the equation x=\frac{-42±6\sqrt{38}}{-18} when ± is plus. Add -42 to 6\sqrt{38}.

$$x=\frac{6\sqrt{38}-42}{-18}$$

Divide -42+6\sqrt{38} by -18.

$$x=\frac{7-\sqrt{38}}{3}$$

Now solve the equation x=\frac{-42±6\sqrt{38}}{-18} when ± is minus. Subtract 6\sqrt{38} from -42.

$$x=\frac{-6\sqrt{38}-42}{-18}$$

Divide -42-6\sqrt{38} by -18.

$$x=\frac{\sqrt{38}+7}{3}$$

The equation is now solved.

$$x=\frac{7-\sqrt{38}}{3} x=\frac{\sqrt{38}+7}{3}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-9x^{2}+42x-11=0$$

Add 11 to both sides. Anything plus zero gives itself.

$$-9x^{2}+42x=11$$

Divide both sides by -9.

$$\frac{-9x^{2}+42x}{-9}=\frac{11}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{42}{-9}x=\frac{11}{-9}$$

Reduce the fraction \frac{42}{-9} to lowest terms by extracting and canceling out 3.

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

Divide 11 by -9.

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

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.

$$x^{2}-\frac{14}{3}x+\left(-\frac{7}{3}\right)^{2}=-\frac{11}{9}+\left(-\frac{7}{3}\right)^{2}$$

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

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

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

$$x^{2}-\frac{14}{3}x+\frac{49}{9}=\frac{38}{9}$$

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}.

$$\left(x-\frac{7}{3}\right)^{2}=\frac{38}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{3}\right)^{2}}=\sqrt{\frac{38}{9}}$$

Simplify.

$$x-\frac{7}{3}=\frac{\sqrt{38}}{3} x-\frac{7}{3}=-\frac{\sqrt{38}}{3}$$

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

$$x=\frac{\sqrt{38}+7}{3} x=\frac{7-\sqrt{38}}{3}$$