Solve 10-9x^2+4x=-6x^2

$$10 - 9 x ^ { 2 } + 4 x = - 6 x ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{34} + 2}{3} \approx 2.610317298$$

$$x=\frac{2-\sqrt{34}}{3}\approx -1.276983965$$

Steps Using the Quadratic Formula

Add 6x^{2} to both sides.

$$10-9x^{2}+4x+6x^{2}=0$$

Combine -9x^{2} and 6x^{2} to get -3x^{2}.

$$10-3x^{2}+4x=0$$

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.

$$-3x^{2}+4x+10=0$$

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-3\right)\times 10}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-4±\sqrt{16+12\times 10}}{2\left(-3\right)}$$

Multiply 12 times 10.

$$x=\frac{-4±\sqrt{16+120}}{2\left(-3\right)}$$

Add 16 to 120.

$$x=\frac{-4±\sqrt{136}}{2\left(-3\right)}$$

Take the square root of 136.

$$x=\frac{-4±2\sqrt{34}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-4±2\sqrt{34}}{-6}$$

Now solve the equation x=\frac{-4±2\sqrt{34}}{-6} when ± is plus. Add -4 to 2\sqrt{34}.

$$x=\frac{2\sqrt{34}-4}{-6}$$

Divide -4+2\sqrt{34} by -6.

$$x=\frac{2-\sqrt{34}}{3}$$

Now solve the equation x=\frac{-4±2\sqrt{34}}{-6} when ± is minus. Subtract 2\sqrt{34} from -4.

$$x=\frac{-2\sqrt{34}-4}{-6}$$

Divide -4-2\sqrt{34} by -6.

$$x=\frac{\sqrt{34}+2}{3}$$

The equation is now solved.

$$x=\frac{2-\sqrt{34}}{3} x=\frac{\sqrt{34}+2}{3}$$

Steps for Completing the Square

Add 6x^{2} to both sides.

$$10-9x^{2}+4x+6x^{2}=0$$

Combine -9x^{2} and 6x^{2} to get -3x^{2}.

$$10-3x^{2}+4x=0$$

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

$$-3x^{2}+4x=-10$$

Divide both sides by -3.

$$\frac{-3x^{2}+4x}{-3}=-\frac{10}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{4}{-3}x=-\frac{10}{-3}$$

Divide 4 by -3.

$$x^{2}-\frac{4}{3}x=-\frac{10}{-3}$$

Divide -10 by -3.

$$x^{2}-\frac{4}{3}x=\frac{10}{3}$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{10}{3}+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{10}{3}+\frac{4}{9}$$

Add \frac{10}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{34}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{34}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{34}{9}}$$

Simplify.

$$x-\frac{2}{3}=\frac{\sqrt{34}}{3} x-\frac{2}{3}=-\frac{\sqrt{34}}{3}$$

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{\sqrt{34}+2}{3} x=\frac{2-\sqrt{34}}{3}$$