Solve x=2x^2+4x+98

$$x = 2 x ^ { 2 } + 4 x + 98$$

Solve for x (complex solution)

$$x=\frac{-5\sqrt{31}i-3}{4}\approx -0.75-6.959705454i$$

$$x=\frac{-3+5\sqrt{31}i}{4}\approx -0.75+6.959705454i$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$x-2x^{2}=4x+98$$

Subtract 4x from both sides.

$$x-2x^{2}-4x=98$$

Combine x and -4x to get -3x.

$$-3x-2x^{2}=98$$

Subtract 98 from both sides.

$$-3x-2x^{2}-98=0$$

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.

$$-2x^{2}-3x-98=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -3 for b, and -98 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-2\right)\left(-98\right)}}{2\left(-2\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-2\right)\left(-98\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8\left(-98\right)}}{2\left(-2\right)}$$

Multiply 8 times -98.

$$x=\frac{-\left(-3\right)±\sqrt{9-784}}{2\left(-2\right)}$$

Add 9 to -784.

$$x=\frac{-\left(-3\right)±\sqrt{-775}}{2\left(-2\right)}$$

Take the square root of -775.

$$x=\frac{-\left(-3\right)±5\sqrt{31}i}{2\left(-2\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±5\sqrt{31}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±5\sqrt{31}i}{-4}$$

Now solve the equation x=\frac{3±5\sqrt{31}i}{-4} when ± is plus. Add 3 to 5i\sqrt{31}.

$$x=\frac{3+5\sqrt{31}i}{-4}$$

Divide 3+5i\sqrt{31} by -4.

$$x=\frac{-5\sqrt{31}i-3}{4}$$

Now solve the equation x=\frac{3±5\sqrt{31}i}{-4} when ± is minus. Subtract 5i\sqrt{31} from 3.

$$x=\frac{-5\sqrt{31}i+3}{-4}$$

Divide 3-5i\sqrt{31} by -4.

$$x=\frac{-3+5\sqrt{31}i}{4}$$

The equation is now solved.

$$x=\frac{-5\sqrt{31}i-3}{4} x=\frac{-3+5\sqrt{31}i}{4}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$x-2x^{2}=4x+98$$

Subtract 4x from both sides.

$$x-2x^{2}-4x=98$$

Combine x and -4x to get -3x.

$$-3x-2x^{2}=98$$

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.

$$-2x^{2}-3x=98$$

Divide both sides by -2.

$$\frac{-2x^{2}-3x}{-2}=\frac{98}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{3}{-2}\right)x=\frac{98}{-2}$$

Divide -3 by -2.

$$x^{2}+\frac{3}{2}x=\frac{98}{-2}$$

Divide 98 by -2.

$$x^{2}+\frac{3}{2}x=-49$$

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.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-49+\left(\frac{3}{4}\right)^{2}$$

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

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-49+\frac{9}{16}$$

Add -49 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{775}{16}$$

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}.

$$\left(x+\frac{3}{4}\right)^{2}=-\frac{775}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{-\frac{775}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{5\sqrt{31}i}{4} x+\frac{3}{4}=-\frac{5\sqrt{31}i}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{-3+5\sqrt{31}i}{4} x=\frac{-5\sqrt{31}i-3}{4}$$