Solve 5=9x^2+2x+3

$$5=9 { x }^{ 2 } +2x+3$$

Solve for x

$$x=\frac{\sqrt{19}-1}{9}\approx 0.373210994$$

$$x=\frac{-\sqrt{19}-1}{9}\approx -0.595433216$$

Steps Using the Quadratic Formula

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

$$9x^{2}+2x+3=5$$

Subtract 5 from both sides.

$$9x^{2}+2x+3-5=0$$

Subtract 5 from 3 to get -2.

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

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

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

Square 2.

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

Multiply -4 times 9.

$$x=\frac{-2±\sqrt{4-36\left(-2\right)}}{2\times 9}$$

Multiply -36 times -2.

$$x=\frac{-2±\sqrt{4+72}}{2\times 9}$$

Add 4 to 72.

$$x=\frac{-2±\sqrt{76}}{2\times 9}$$

Take the square root of 76.

$$x=\frac{-2±2\sqrt{19}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-2±2\sqrt{19}}{18}$$

Now solve the equation x=\frac{-2±2\sqrt{19}}{18} when ± is plus. Add -2 to 2\sqrt{19}.

$$x=\frac{2\sqrt{19}-2}{18}$$

Divide -2+2\sqrt{19} by 18.

$$x=\frac{\sqrt{19}-1}{9}$$

Now solve the equation x=\frac{-2±2\sqrt{19}}{18} when ± is minus. Subtract 2\sqrt{19} from -2.

$$x=\frac{-2\sqrt{19}-2}{18}$$

Divide -2-2\sqrt{19} by 18.

$$x=\frac{-\sqrt{19}-1}{9}$$

The equation is now solved.

$$x=\frac{\sqrt{19}-1}{9} x=\frac{-\sqrt{19}-1}{9}$$

Steps for Completing the Square

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

$$9x^{2}+2x+3=5$$

Subtract 3 from both sides.

$$9x^{2}+2x=5-3$$

Subtract 3 from 5 to get 2.

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

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

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

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

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

Square \frac{1}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{9}x+\frac{1}{81}=\frac{2}{9}+\frac{1}{81}$$

Add \frac{2}{9} to \frac{1}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{9}x+\frac{1}{81}=\frac{19}{81}$$

Factor x^{2}+\frac{2}{9}x+\frac{1}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{9}\right)^{2}=\frac{19}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{9}\right)^{2}}=\sqrt{\frac{19}{81}}$$

Simplify.

$$x+\frac{1}{9}=\frac{\sqrt{19}}{9} x+\frac{1}{9}=-\frac{\sqrt{19}}{9}$$

Subtract \frac{1}{9} from both sides of the equation.

$$x=\frac{\sqrt{19}-1}{9} x=\frac{-\sqrt{19}-1}{9}$$