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

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{35}i+5}{6}\approx 0.833333333-0.986013297i$$

$$x=\frac{5+\sqrt{35}i}{6}\approx 0.833333333+0.986013297i$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

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

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.

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

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

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

Square 5.

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

Multiply -4 times -3.

$$x=\frac{-5±\sqrt{25+12\left(-5\right)}}{2\left(-3\right)}$$

Multiply 12 times -5.

$$x=\frac{-5±\sqrt{25-60}}{2\left(-3\right)}$$

Add 25 to -60.

$$x=\frac{-5±\sqrt{-35}}{2\left(-3\right)}$$

Take the square root of -35.

$$x=\frac{-5±\sqrt{35}i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-5±\sqrt{35}i}{-6}$$

Now solve the equation x=\frac{-5±\sqrt{35}i}{-6} when ± is plus. Add -5 to i\sqrt{35}.

$$x=\frac{-5+\sqrt{35}i}{-6}$$

Divide -5+i\sqrt{35} by -6.

$$x=\frac{-\sqrt{35}i+5}{6}$$

Now solve the equation x=\frac{-5±\sqrt{35}i}{-6} when ± is minus. Subtract i\sqrt{35} from -5.

$$x=\frac{-\sqrt{35}i-5}{-6}$$

Divide -5-i\sqrt{35} by -6.

$$x=\frac{5+\sqrt{35}i}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{35}i+5}{6} x=\frac{5+\sqrt{35}i}{6}$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

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

Add 2x to both sides.

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

Combine 3x and 2x to get 5x.

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

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

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 5 by -3.

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

Divide 5 by -3.

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

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{5}{3}+\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{5}{3}+\frac{25}{36}$$

Add -\frac{5}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{35}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{6}\right)^{2}=-\frac{35}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{-\frac{35}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{\sqrt{35}i}{6} x-\frac{5}{6}=-\frac{\sqrt{35}i}{6}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{5+\sqrt{35}i}{6} x=\frac{-\sqrt{35}i+5}{6}$$