Solve 2x^2=10x+3

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

Solve for x

$$x = \frac{\sqrt{31} + 5}{2} \approx 5.283882181$$

$$x=\frac{5-\sqrt{31}}{2}\approx -0.283882181$$

Steps Using the Quadratic Formula

Subtract 10x from both sides.

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

Subtract 3 from both sides.

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

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

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

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-10\right)±\sqrt{100-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-\left(-10\right)±\sqrt{100+24}}{2\times 2}$$

Add 100 to 24.

$$x=\frac{-\left(-10\right)±\sqrt{124}}{2\times 2}$$

Take the square root of 124.

$$x=\frac{-\left(-10\right)±2\sqrt{31}}{2\times 2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{31}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{10±2\sqrt{31}}{4}$$

Now solve the equation x=\frac{10±2\sqrt{31}}{4} when ± is plus. Add 10 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}+10}{4}$$

Divide 10+2\sqrt{31} by 4.

$$x=\frac{\sqrt{31}+5}{2}$$

Now solve the equation x=\frac{10±2\sqrt{31}}{4} when ± is minus. Subtract 2\sqrt{31} from 10.

$$x=\frac{10-2\sqrt{31}}{4}$$

Divide 10-2\sqrt{31} by 4.

$$x=\frac{5-\sqrt{31}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{31}+5}{2} x=\frac{5-\sqrt{31}}{2}$$

Steps for Completing the Square

Subtract 10x from both sides.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{10}{2}\right)x=\frac{3}{2}$$

Divide -10 by 2.

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

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{3}{2}+\frac{25}{4}$$

Add \frac{3}{2} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=\frac{31}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{31}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{31}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{31}}{2} x-\frac{5}{2}=-\frac{\sqrt{31}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{31}+5}{2} x=\frac{5-\sqrt{31}}{2}$$