Solve 2x^2=3x+10

$$2 x ^ { 2 } = 3 x + 10$$

Solve for x

$$x = \frac{\sqrt{89} + 3}{4} \approx 3.108495283$$

$$x=\frac{3-\sqrt{89}}{4}\approx -1.608495283$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$2x^{2}-3x=10$$

Subtract 10 from both sides.

$$2x^{2}-3x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-10\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-10\right)}}{2\times 2}$$

Multiply -8 times -10.

$$x=\frac{-\left(-3\right)±\sqrt{9+80}}{2\times 2}$$

Add 9 to 80.

$$x=\frac{-\left(-3\right)±\sqrt{89}}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{89}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±\sqrt{89}}{4}$$

Now solve the equation x=\frac{3±\sqrt{89}}{4} when ± is plus. Add 3 to \sqrt{89}.

$$x=\frac{\sqrt{89}+3}{4}$$

Now solve the equation x=\frac{3±\sqrt{89}}{4} when ± is minus. Subtract \sqrt{89} from 3.

$$x=\frac{3-\sqrt{89}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{89}+3}{4} x=\frac{3-\sqrt{89}}{4}$$

Steps for Completing the Square

Subtract 3x from both sides.

$$2x^{2}-3x=10$$

Divide both sides by 2.

$$\frac{2x^{2}-3x}{2}=\frac{10}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{3}{2}x=\frac{10}{2}$$

Divide 10 by 2.

$$x^{2}-\frac{3}{2}x=5$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=5+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=5+\frac{9}{16}$$

Add 5 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{89}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{89}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{89}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{89}}{4} x-\frac{3}{4}=-\frac{\sqrt{89}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{89}+3}{4} x=\frac{3-\sqrt{89}}{4}$$