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

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{11}i-3}{2}\approx -1.5-1.658312395i$$

$$x=\frac{-3+\sqrt{11}i}{2}\approx -1.5+1.658312395i$$

Steps Using the Quadratic Formula

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\left(-5\right)}}{2\left(-1\right)}$$

Multiply 4 times -5.

$$x=\frac{-\left(-3\right)±\sqrt{9-20}}{2\left(-1\right)}$$

Add 9 to -20.

$$x=\frac{-\left(-3\right)±\sqrt{-11}}{2\left(-1\right)}$$

Take the square root of -11.

$$x=\frac{-\left(-3\right)±\sqrt{11}i}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{11}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±\sqrt{11}i}{-2}$$

Now solve the equation x=\frac{3±\sqrt{11}i}{-2} when ± is plus. Add 3 to i\sqrt{11}.

$$x=\frac{3+\sqrt{11}i}{-2}$$

Divide 3+i\sqrt{11} by -2.

$$x=\frac{-\sqrt{11}i-3}{2}$$

Now solve the equation x=\frac{3±\sqrt{11}i}{-2} when ± is minus. Subtract i\sqrt{11} from 3.

$$x=\frac{-\sqrt{11}i+3}{-2}$$

Divide 3-i\sqrt{11} by -2.

$$x=\frac{-3+\sqrt{11}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{11}i-3}{2} x=\frac{-3+\sqrt{11}i}{2}$$

Steps for Completing the Square

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

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

Add 5 to both sides. Anything plus zero gives itself.

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

Divide both sides by -1.

$$\frac{-x^{2}-3x}{-1}=\frac{5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{3}{-1}\right)x=\frac{5}{-1}$$

Divide -3 by -1.

$$x^{2}+3x=\frac{5}{-1}$$

Divide 5 by -1.

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-5+\frac{9}{4}$$

Add -5 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=-\frac{11}{4}$$

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

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{11}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{11}i}{2} x+\frac{3}{2}=-\frac{\sqrt{11}i}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{-3+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-3}{2}$$