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

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

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{11}i}{4}\approx -0.25+0.829156198i$$

$$x=\frac{-\sqrt{11}i-1}{4}\approx -0.25-0.829156198i$$

Steps Using the Quadratic Formula

Subtract \frac{x^{2}-3}{5x+2} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{5x+2}{5x+2}.

$$\frac{x\left(5x+2\right)}{5x+2}-\frac{x^{2}-3}{5x+2}=0$$

Since \frac{x\left(5x+2\right)}{5x+2} and \frac{x^{2}-3}{5x+2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(5x+2\right)-\left(x^{2}-3\right)}{5x+2}=0$$

Do the multiplications in x\left(5x+2\right)-\left(x^{2}-3\right).

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

Combine like terms in 5x^{2}+2x-x^{2}+3.

$$\frac{4x^{2}+2x+3}{5x+2}=0$$

Variable x cannot be equal to -\frac{2}{5} since division by zero is not defined. Multiply both sides of the equation by 5x+2.

$$4x^{2}+2x+3=0$$

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

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\times 3}}{2\times 4}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-2±\sqrt{4-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$x=\frac{-2±\sqrt{4-48}}{2\times 4}$$

Add 4 to -48.

$$x=\frac{-2±\sqrt{-44}}{2\times 4}$$

Take the square root of -44.

$$x=\frac{-2±2\sqrt{11}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±2\sqrt{11}i}{8}$$

Now solve the equation x=\frac{-2±2\sqrt{11}i}{8} when ± is plus. Add -2 to 2i\sqrt{11}.

$$x=\frac{-2+2\sqrt{11}i}{8}$$

Divide -2+2i\sqrt{11} by 8.

$$x=\frac{-1+\sqrt{11}i}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{11}i}{8} when ± is minus. Subtract 2i\sqrt{11} from -2.

$$x=\frac{-2\sqrt{11}i-2}{8}$$

Divide -2-2i\sqrt{11} by 8.

$$x=\frac{-\sqrt{11}i-1}{4}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{11}i}{4} x=\frac{-\sqrt{11}i-1}{4}$$

Steps for Completing the Square

Subtract \frac{x^{2}-3}{5x+2} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{5x+2}{5x+2}.

$$\frac{x\left(5x+2\right)}{5x+2}-\frac{x^{2}-3}{5x+2}=0$$

Since \frac{x\left(5x+2\right)}{5x+2} and \frac{x^{2}-3}{5x+2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(5x+2\right)-\left(x^{2}-3\right)}{5x+2}=0$$

Do the multiplications in x\left(5x+2\right)-\left(x^{2}-3\right).

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

Combine like terms in 5x^{2}+2x-x^{2}+3.

$$\frac{4x^{2}+2x+3}{5x+2}=0$$

Variable x cannot be equal to -\frac{2}{5} since division by zero is not defined. Multiply both sides of the equation by 5x+2.

$$4x^{2}+2x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+2x=-3$$

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{3}{4}+\frac{1}{16}$$

Add -\frac{3}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{11}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{11}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{11}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{11}i}{4} x+\frac{1}{4}=-\frac{\sqrt{11}i}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{-1+\sqrt{11}i}{4} x=\frac{-\sqrt{11}i-1}{4}$$