Solve 5x+2=2x^2

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

Solve for x

$$x = \frac{\sqrt{41} + 5}{4} \approx 2.850781059$$

$$x=\frac{5-\sqrt{41}}{4}\approx -0.350781059$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-5±\sqrt{25+8\times 2}}{2\left(-2\right)}$$

Multiply 8 times 2.

$$x=\frac{-5±\sqrt{25+16}}{2\left(-2\right)}$$

Add 25 to 16.

$$x=\frac{-5±\sqrt{41}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-5±\sqrt{41}}{-4}$$

Now solve the equation x=\frac{-5±\sqrt{41}}{-4} when ± is plus. Add -5 to \sqrt{41}.

$$x=\frac{\sqrt{41}-5}{-4}$$

Divide -5+\sqrt{41} by -4.

$$x=\frac{5-\sqrt{41}}{4}$$

Now solve the equation x=\frac{-5±\sqrt{41}}{-4} when ± is minus. Subtract \sqrt{41} from -5.

$$x=\frac{-\sqrt{41}-5}{-4}$$

Divide -5-\sqrt{41} by -4.

$$x=\frac{\sqrt{41}+5}{4}$$

The equation is now solved.

$$x=\frac{5-\sqrt{41}}{4} x=\frac{\sqrt{41}+5}{4}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$5x-2x^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 5 by -2.

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

Divide -2 by -2.

$$x^{2}-\frac{5}{2}x=1$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=1+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=1+\frac{25}{16}$$

Add 1 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{41}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{41}}{4} x-\frac{5}{4}=-\frac{\sqrt{41}}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{\sqrt{41}+5}{4} x=\frac{5-\sqrt{41}}{4}$$