Solve x=5-9*x^2

$$x = 5-9 \times { x }^{ 2 }$$

Solve for x

$$x=\frac{\sqrt{181}-1}{18}\approx 0.691868003$$

$$x=\frac{-\sqrt{181}-1}{18}\approx -0.802979114$$

Steps Using the Quadratic Formula

Subtract 5 from both sides.

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

Add 9x^{2} to both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 1 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 9\left(-5\right)}}{2\times 9}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 9\left(-5\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-1±\sqrt{1-36\left(-5\right)}}{2\times 9}$$

Multiply -36 times -5.

$$x=\frac{-1±\sqrt{1+180}}{2\times 9}$$

Add 1 to 180.

$$x=\frac{-1±\sqrt{181}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-1±\sqrt{181}}{18}$$

Now solve the equation x=\frac{-1±\sqrt{181}}{18} when ± is plus. Add -1 to \sqrt{181}.

$$x=\frac{\sqrt{181}-1}{18}$$

Now solve the equation x=\frac{-1±\sqrt{181}}{18} when ± is minus. Subtract \sqrt{181} from -1.

$$x=\frac{-\sqrt{181}-1}{18}$$

The equation is now solved.

$$x=\frac{\sqrt{181}-1}{18} x=\frac{-\sqrt{181}-1}{18}$$

Steps for Completing the Square

Add 9x^{2} to both sides.

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

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.

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

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{1}{9}x=\frac{5}{9}$$

Divide \frac{1}{9}, the coefficient of the x term, by 2 to get \frac{1}{18}. Then add the square of \frac{1}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{9}x+\left(\frac{1}{18}\right)^{2}=\frac{5}{9}+\left(\frac{1}{18}\right)^{2}$$

Square \frac{1}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{9}x+\frac{1}{324}=\frac{5}{9}+\frac{1}{324}$$

Add \frac{5}{9} to \frac{1}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{9}x+\frac{1}{324}=\frac{181}{324}$$

Factor x^{2}+\frac{1}{9}x+\frac{1}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{18}\right)^{2}=\frac{181}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{18}\right)^{2}}=\sqrt{\frac{181}{324}}$$

Simplify.

$$x+\frac{1}{18}=\frac{\sqrt{181}}{18} x+\frac{1}{18}=-\frac{\sqrt{181}}{18}$$

Subtract \frac{1}{18} from both sides of the equation.

$$x=\frac{\sqrt{181}-1}{18} x=\frac{-\sqrt{181}-1}{18}$$