Solve 2=2x^2+7x+9/2

$$2 = 2 x ^ { 2 } + 7 x + \frac { 9 } { 2 }$$

Solve for x

$$x=\frac{\sqrt{29}-7}{4}\approx -0.403708798$$

$$x=\frac{-\sqrt{29}-7}{4}\approx -3.096291202$$

Steps Using the Quadratic Formula

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

$$2x^{2}+7x+\frac{9}{2}=2$$

Subtract 2 from both sides.

$$2x^{2}+7x+\frac{9}{2}-2=0$$

Subtract 2 from \frac{9}{2} to get \frac{5}{2}.

$$2x^{2}+7x+\frac{5}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and \frac{5}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times \frac{5}{2}}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times \frac{5}{2}}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times \frac{5}{2}}}{2\times 2}$$

Multiply -8 times \frac{5}{2}.

$$x=\frac{-7±\sqrt{49-20}}{2\times 2}$$

Add 49 to -20.

$$x=\frac{-7±\sqrt{29}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±\sqrt{29}}{4}$$

Now solve the equation x=\frac{-7±\sqrt{29}}{4} when ± is plus. Add -7 to \sqrt{29}.

$$x=\frac{\sqrt{29}-7}{4}$$

Now solve the equation x=\frac{-7±\sqrt{29}}{4} when ± is minus. Subtract \sqrt{29} from -7.

$$x=\frac{-\sqrt{29}-7}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{29}-7}{4} x=\frac{-\sqrt{29}-7}{4}$$

Steps for Completing the Square

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

$$2x^{2}+7x+\frac{9}{2}=2$$

Subtract \frac{9}{2} from both sides.

$$2x^{2}+7x=2-\frac{9}{2}$$

Subtract \frac{9}{2} from 2 to get -\frac{5}{2}.

$$2x^{2}+7x=-\frac{5}{2}$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=-\frac{\frac{5}{2}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=-\frac{\frac{5}{2}}{2}$$

Divide -\frac{5}{2} by 2.

$$x^{2}+\frac{7}{2}x=-\frac{5}{4}$$

Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=-\frac{5}{4}+\left(\frac{7}{4}\right)^{2}$$

Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-\frac{5}{4}+\frac{49}{16}$$

Add -\frac{5}{4} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{29}{16}$$

Factor x^{2}+\frac{7}{2}x+\frac{49}{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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{29}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{29}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{\sqrt{29}}{4} x+\frac{7}{4}=-\frac{\sqrt{29}}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=\frac{\sqrt{29}-7}{4} x=\frac{-\sqrt{29}-7}{4}$$