Solve 10-6x-9x^2=-9x

$$10 - 6 x - 9 x ^ { 2 } = - 9 x$$

Solve for x

$$x = \frac{\sqrt{41} + 1}{6} \approx 1.23385404$$

$$x=\frac{1-\sqrt{41}}{6}\approx -0.900520706$$

Steps Using the Quadratic Formula

Add 9x to both sides.

$$10-6x-9x^{2}+9x=0$$

Combine -6x and 9x to get 3x.

$$10+3x-9x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-9x^{2}+3x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 3 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-9\right)\times 10}}{2\left(-9\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-9\right)\times 10}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-3±\sqrt{9+36\times 10}}{2\left(-9\right)}$$

Multiply 36 times 10.

$$x=\frac{-3±\sqrt{9+360}}{2\left(-9\right)}$$

Add 9 to 360.

$$x=\frac{-3±\sqrt{369}}{2\left(-9\right)}$$

Take the square root of 369.

$$x=\frac{-3±3\sqrt{41}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-3±3\sqrt{41}}{-18}$$

Now solve the equation x=\frac{-3±3\sqrt{41}}{-18} when ± is plus. Add -3 to 3\sqrt{41}.

$$x=\frac{3\sqrt{41}-3}{-18}$$

Divide -3+3\sqrt{41} by -18.

$$x=\frac{1-\sqrt{41}}{6}$$

Now solve the equation x=\frac{-3±3\sqrt{41}}{-18} when ± is minus. Subtract 3\sqrt{41} from -3.

$$x=\frac{-3\sqrt{41}-3}{-18}$$

Divide -3-3\sqrt{41} by -18.

$$x=\frac{\sqrt{41}+1}{6}$$

The equation is now solved.

$$x=\frac{1-\sqrt{41}}{6} x=\frac{\sqrt{41}+1}{6}$$

Steps for Completing the Square

Add 9x to both sides.

$$10-6x-9x^{2}+9x=0$$

Combine -6x and 9x to get 3x.

$$10+3x-9x^{2}=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$3x-9x^{2}=-10$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-9x^{2}+3x=-10$$

Divide both sides by -9.

$$\frac{-9x^{2}+3x}{-9}=-\frac{10}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{3}{-9}x=-\frac{10}{-9}$$

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

$$x^{2}-\frac{1}{3}x=-\frac{10}{-9}$$

Divide -10 by -9.

$$x^{2}-\frac{1}{3}x=\frac{10}{9}$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{10}{9}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{10}{9}+\frac{1}{36}$$

Add \frac{10}{9} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{41}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{41}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{41}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{41}}{6} x-\frac{1}{6}=-\frac{\sqrt{41}}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{\sqrt{41}+1}{6} x=\frac{1-\sqrt{41}}{6}$$