Solve x=5x^2+6x+3/x+1

$$x = \frac { 5 x ^ { 2 } + 6 x + 3 } { x + 1 }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{23}i-5}{8}\approx -0.625-0.59947894i$$

$$x=\frac{-5+\sqrt{23}i}{8}\approx -0.625+0.59947894i$$

Steps Using the Quadratic Formula

Subtract \frac{5x^{2}+6x+3}{x+1} from both sides.

$$x-\frac{5x^{2}+6x+3}{x+1}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+1}{x+1}.

$$\frac{x\left(x+1\right)}{x+1}-\frac{5x^{2}+6x+3}{x+1}=0$$

Since \frac{x\left(x+1\right)}{x+1} and \frac{5x^{2}+6x+3}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+1\right)-\left(5x^{2}+6x+3\right)}{x+1}=0$$

Do the multiplications in x\left(x+1\right)-\left(5x^{2}+6x+3\right).

$$\frac{x^{2}+x-5x^{2}-6x-3}{x+1}=0$$

Combine like terms in x^{2}+x-5x^{2}-6x-3.

$$\frac{-4x^{2}-5x-3}{x+1}=0$$

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$-4x^{2}-5x-3=0$$

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

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

Square -5.

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

Multiply -4 times -4.

$$x=\frac{-\left(-5\right)±\sqrt{25+16\left(-3\right)}}{2\left(-4\right)}$$

Multiply 16 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25-48}}{2\left(-4\right)}$$

Add 25 to -48.

$$x=\frac{-\left(-5\right)±\sqrt{-23}}{2\left(-4\right)}$$

Take the square root of -23.

$$x=\frac{-\left(-5\right)±\sqrt{23}i}{2\left(-4\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{23}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{5±\sqrt{23}i}{-8}$$

Now solve the equation x=\frac{5±\sqrt{23}i}{-8} when ± is plus. Add 5 to i\sqrt{23}.

$$x=\frac{5+\sqrt{23}i}{-8}$$

Divide 5+i\sqrt{23} by -8.

$$x=\frac{-\sqrt{23}i-5}{8}$$

Now solve the equation x=\frac{5±\sqrt{23}i}{-8} when ± is minus. Subtract i\sqrt{23} from 5.

$$x=\frac{-\sqrt{23}i+5}{-8}$$

Divide 5-i\sqrt{23} by -8.

$$x=\frac{-5+\sqrt{23}i}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{23}i-5}{8} x=\frac{-5+\sqrt{23}i}{8}$$

Steps for Completing the Square

Subtract \frac{5x^{2}+6x+3}{x+1} from both sides.

$$x-\frac{5x^{2}+6x+3}{x+1}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+1}{x+1}.

$$\frac{x\left(x+1\right)}{x+1}-\frac{5x^{2}+6x+3}{x+1}=0$$

Since \frac{x\left(x+1\right)}{x+1} and \frac{5x^{2}+6x+3}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+1\right)-\left(5x^{2}+6x+3\right)}{x+1}=0$$

Do the multiplications in x\left(x+1\right)-\left(5x^{2}+6x+3\right).

$$\frac{x^{2}+x-5x^{2}-6x-3}{x+1}=0$$

Combine like terms in x^{2}+x-5x^{2}-6x-3.

$$\frac{-4x^{2}-5x-3}{x+1}=0$$

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$-4x^{2}-5x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$-4x^{2}-5x=3$$

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide -5 by -4.

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

Divide 3 by -4.

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

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=-\frac{3}{4}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-\frac{3}{4}+\frac{25}{64}$$

Add -\frac{3}{4} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-\frac{23}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=-\frac{23}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{-\frac{23}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{\sqrt{23}i}{8} x+\frac{5}{8}=-\frac{\sqrt{23}i}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{-5+\sqrt{23}i}{8} x=\frac{-\sqrt{23}i-5}{8}$$