Solve 2x^2-5x^2-2x-3=0

$$2 x ^ { 2 } - 5 x ^ { 2 } - 2 x - 3 = 0$$

Solve for x (complex solution)

$$x=\frac{-2\sqrt{2}i-1}{3}\approx -0.333333333-0.942809042i$$

$$x=\frac{-1+2\sqrt{2}i}{3}\approx -0.333333333+0.942809042i$$

Steps Using the Quadratic Formula

Combine 2x^{2} and -5x^{2} to get -3x^{2}.

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

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

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

Square -2.

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

Multiply -4 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12\left(-3\right)}}{2\left(-3\right)}$$

Multiply 12 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4-36}}{2\left(-3\right)}$$

Add 4 to -36.

$$x=\frac{-\left(-2\right)±\sqrt{-32}}{2\left(-3\right)}$$

Take the square root of -32.

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

The opposite of -2 is 2.

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

Multiply 2 times -3.

$$x=\frac{2±4\sqrt{2}i}{-6}$$

Now solve the equation x=\frac{2±4\sqrt{2}i}{-6} when ± is plus. Add 2 to 4i\sqrt{2}.

$$x=\frac{2+4\sqrt{2}i}{-6}$$

Divide 2+4i\sqrt{2} by -6.

$$x=\frac{-2\sqrt{2}i-1}{3}$$

Now solve the equation x=\frac{2±4\sqrt{2}i}{-6} when ± is minus. Subtract 4i\sqrt{2} from 2.

$$x=\frac{-4\sqrt{2}i+2}{-6}$$

Divide 2-4i\sqrt{2} by -6.

$$x=\frac{-1+2\sqrt{2}i}{3}$$

The equation is now solved.

$$x=\frac{-2\sqrt{2}i-1}{3} x=\frac{-1+2\sqrt{2}i}{3}$$

Steps for Completing the Square

Combine 2x^{2} and -5x^{2} to get -3x^{2}.

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

Add 3 to both sides. Anything plus zero gives itself.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -2 by -3.

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

Divide 3 by -3.

$$x^{2}+\frac{2}{3}x=-1$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-1+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-1+\frac{1}{9}$$

Add -1 to \frac{1}{9}.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{8}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{8}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{8}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{2\sqrt{2}i}{3} x+\frac{1}{3}=-\frac{2\sqrt{2}i}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{-1+2\sqrt{2}i}{3} x=\frac{-2\sqrt{2}i-1}{3}$$