Solve 65=2*x^2+9*x+5=

$$65=2 \times { x }^{ 2 } +9 \times x+5=$$

Solve for x

$$x = \frac{\sqrt{561} - 9}{4} \approx 3.671359641$$

$$x=\frac{-\sqrt{561}-9}{4}\approx -8.171359641$$

Steps Using the Quadratic Formula

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

$$2x^{2}+9x+5=65$$

Subtract 65 from both sides.

$$2x^{2}+9x+5-65=0$$

Subtract 65 from 5 to get -60.

$$2x^{2}+9x-60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 9 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 2\left(-60\right)}}{2\times 2}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 2\left(-60\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-9±\sqrt{81-8\left(-60\right)}}{2\times 2}$$

Multiply -8 times -60.

$$x=\frac{-9±\sqrt{81+480}}{2\times 2}$$

Add 81 to 480.

$$x=\frac{-9±\sqrt{561}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-9±\sqrt{561}}{4}$$

Now solve the equation x=\frac{-9±\sqrt{561}}{4} when ± is plus. Add -9 to \sqrt{561}.

$$x=\frac{\sqrt{561}-9}{4}$$

Now solve the equation x=\frac{-9±\sqrt{561}}{4} when ± is minus. Subtract \sqrt{561} from -9.

$$x=\frac{-\sqrt{561}-9}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{561}-9}{4} x=\frac{-\sqrt{561}-9}{4}$$

Steps for Completing the Square

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

$$2x^{2}+9x+5=65$$

Subtract 5 from both sides.

$$2x^{2}+9x=65-5$$

Subtract 5 from 65 to get 60.

$$2x^{2}+9x=60$$

Divide both sides by 2.

$$\frac{2x^{2}+9x}{2}=\frac{60}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{9}{2}x=\frac{60}{2}$$

Divide 60 by 2.

$$x^{2}+\frac{9}{2}x=30$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{2}x+\left(\frac{9}{4}\right)^{2}=30+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=30+\frac{81}{16}$$

Add 30 to \frac{81}{16}.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{561}{16}$$

Factor x^{2}+\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{4}\right)^{2}=\frac{561}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{\frac{561}{16}}$$

Simplify.

$$x+\frac{9}{4}=\frac{\sqrt{561}}{4} x+\frac{9}{4}=-\frac{\sqrt{561}}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$x=\frac{\sqrt{561}-9}{4} x=\frac{-\sqrt{561}-9}{4}$$