Solve x=-5x^2-5x+6

$$x=-5 { x }^{ 2 } -5x+6$$

Solve for x

$$x=\frac{\sqrt{39}-3}{5}\approx 0.6489996$$

$$x=\frac{-\sqrt{39}-3}{5}\approx -1.8489996$$

Steps Using the Quadratic Formula

Add 5x^{2} to both sides.

$$x+5x^{2}=-5x+6$$

Add 5x to both sides.

$$x+5x^{2}+5x=6$$

Combine x and 5x to get 6x.

$$6x+5x^{2}=6$$

Subtract 6 from both sides.

$$6x+5x^{2}-6=0$$

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.

$$5x^{2}+6x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 6 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\left(-6\right)}}{2\times 5}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 5\left(-6\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-6±\sqrt{36-20\left(-6\right)}}{2\times 5}$$

Multiply -20 times -6.

$$x=\frac{-6±\sqrt{36+120}}{2\times 5}$$

Add 36 to 120.

$$x=\frac{-6±\sqrt{156}}{2\times 5}$$

Take the square root of 156.

$$x=\frac{-6±2\sqrt{39}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-6±2\sqrt{39}}{10}$$

Now solve the equation x=\frac{-6±2\sqrt{39}}{10} when ± is plus. Add -6 to 2\sqrt{39}.

$$x=\frac{2\sqrt{39}-6}{10}$$

Divide -6+2\sqrt{39} by 10.

$$x=\frac{\sqrt{39}-3}{5}$$

Now solve the equation x=\frac{-6±2\sqrt{39}}{10} when ± is minus. Subtract 2\sqrt{39} from -6.

$$x=\frac{-2\sqrt{39}-6}{10}$$

Divide -6-2\sqrt{39} by 10.

$$x=\frac{-\sqrt{39}-3}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{39}-3}{5} x=\frac{-\sqrt{39}-3}{5}$$

Steps for Completing the Square

Add 5x^{2} to both sides.

$$x+5x^{2}=-5x+6$$

Add 5x to both sides.

$$x+5x^{2}+5x=6$$

Combine x and 5x to get 6x.

$$6x+5x^{2}=6$$

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.

$$5x^{2}+6x=6$$

Divide both sides by 5.

$$\frac{5x^{2}+6x}{5}=\frac{6}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{6}{5}x=\frac{6}{5}$$

Divide \frac{6}{5}, the coefficient of the x term, by 2 to get \frac{3}{5}. Then add the square of \frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=\frac{6}{5}+\left(\frac{3}{5}\right)^{2}$$

Square \frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{6}{5}+\frac{9}{25}$$

Add \frac{6}{5} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{39}{25}$$

Factor x^{2}+\frac{6}{5}x+\frac{9}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{5}\right)^{2}=\frac{39}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{39}{25}}$$

Simplify.

$$x+\frac{3}{5}=\frac{\sqrt{39}}{5} x+\frac{3}{5}=-\frac{\sqrt{39}}{5}$$

Subtract \frac{3}{5} from both sides of the equation.

$$x=\frac{\sqrt{39}-3}{5} x=\frac{-\sqrt{39}-3}{5}$$