Solve (x)=3x^2-4x+11/5

$$( x ) = 3 x ^ { 2 } - 4 x + \frac { 11 } { 5 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt{35}i}{30}+\frac{5}{6}\approx 0.833333333+0.197202659i$$

$$x=-\frac{\sqrt{35}i}{30}+\frac{5}{6}\approx 0.833333333-0.197202659i$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$x-3x^{2}=-4x+\frac{11}{5}$$

Add 4x to both sides.

$$x-3x^{2}+4x=\frac{11}{5}$$

Combine x and 4x to get 5x.

$$5x-3x^{2}=\frac{11}{5}$$

Subtract \frac{11}{5} from both sides.

$$5x-3x^{2}-\frac{11}{5}=0$$

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.

$$-3x^{2}+5x-\frac{11}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 5 for b, and -\frac{11}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-3\right)\left(-\frac{11}{5}\right)}}{2\left(-3\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-3\right)\left(-\frac{11}{5}\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-5±\sqrt{25+12\left(-\frac{11}{5}\right)}}{2\left(-3\right)}$$

Multiply 12 times -\frac{11}{5}.

$$x=\frac{-5±\sqrt{25-\frac{132}{5}}}{2\left(-3\right)}$$

Add 25 to -\frac{132}{5}.

$$x=\frac{-5±\sqrt{-\frac{7}{5}}}{2\left(-3\right)}$$

Take the square root of -\frac{7}{5}.

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

Multiply 2 times -3.

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

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

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

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

$$x=-\frac{\sqrt{35}i}{30}+\frac{5}{6}$$

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

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

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

$$x=\frac{\sqrt{35}i}{30}+\frac{5}{6}$$

The equation is now solved.

$$x=-\frac{\sqrt{35}i}{30}+\frac{5}{6} x=\frac{\sqrt{35}i}{30}+\frac{5}{6}$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$x-3x^{2}=-4x+\frac{11}{5}$$

Add 4x to both sides.

$$x-3x^{2}+4x=\frac{11}{5}$$

Combine x and 4x to get 5x.

$$5x-3x^{2}=\frac{11}{5}$$

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.

$$-3x^{2}+5x=\frac{11}{5}$$

Divide both sides by -3.

$$\frac{-3x^{2}+5x}{-3}=\frac{\frac{11}{5}}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{5}{-3}x=\frac{\frac{11}{5}}{-3}$$

Divide 5 by -3.

$$x^{2}-\frac{5}{3}x=\frac{\frac{11}{5}}{-3}$$

Divide \frac{11}{5} by -3.

$$x^{2}-\frac{5}{3}x=-\frac{11}{15}$$

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.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{11}{15}+\left(-\frac{5}{6}\right)^{2}$$

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

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{11}{15}+\frac{25}{36}$$

Add -\frac{11}{15} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{7}{180}$$

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}.

$$\left(x-\frac{5}{6}\right)^{2}=-\frac{7}{180}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{-\frac{7}{180}}$$

Simplify.

$$x-\frac{5}{6}=\frac{\sqrt{35}i}{30} x-\frac{5}{6}=-\frac{\sqrt{35}i}{30}$$

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

$$x=\frac{\sqrt{35}i}{30}+\frac{5}{6} x=-\frac{\sqrt{35}i}{30}+\frac{5}{6}$$