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

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

Solve for x (complex solution)

$$x=-\frac{\sqrt{35}i}{10}+\frac{1}{2}\approx 0.5-0.591607978i$$

$$x=\frac{\sqrt{35}i}{10}+\frac{1}{2}\approx 0.5+0.591607978i$$

Steps Using the Quadratic Formula

Subtract 3 from both sides.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

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

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

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

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-5±\sqrt{25+20\left(-3\right)}}{2\left(-5\right)}$$

Multiply 20 times -3.

$$x=\frac{-5±\sqrt{25-60}}{2\left(-5\right)}$$

Add 25 to -60.

$$x=\frac{-5±\sqrt{-35}}{2\left(-5\right)}$$

Take the square root of -35.

$$x=\frac{-5±\sqrt{35}i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-5±\sqrt{35}i}{-10}$$

Now solve the equation x=\frac{-5±\sqrt{35}i}{-10} when ± is plus. Add -5 to i\sqrt{35}.

$$x=\frac{-5+\sqrt{35}i}{-10}$$

Divide -5+i\sqrt{35} by -10.

$$x=-\frac{\sqrt{35}i}{10}+\frac{1}{2}$$

Now solve the equation x=\frac{-5±\sqrt{35}i}{-10} when ± is minus. Subtract i\sqrt{35} from -5.

$$x=\frac{-\sqrt{35}i-5}{-10}$$

Divide -5-i\sqrt{35} by -10.

$$x=\frac{\sqrt{35}i}{10}+\frac{1}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{35}i}{10}+\frac{1}{2} x=\frac{\sqrt{35}i}{10}+\frac{1}{2}$$

Steps for Completing the Square

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

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

Divide both sides by -5.

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

Dividing by -5 undoes the multiplication by -5.

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

Divide 5 by -5.

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

Divide 3 by -5.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-\frac{3}{5}+\frac{1}{4}$$

Add -\frac{3}{5} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=-\frac{7}{20}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{7}{20}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{7}{20}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{35}i}{10} x-\frac{1}{2}=-\frac{\sqrt{35}i}{10}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{35}i}{10}+\frac{1}{2} x=-\frac{\sqrt{35}i}{10}+\frac{1}{2}$$