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

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

Solve for x

$$x = \frac{\sqrt{33} + 7}{4} \approx 3.186140662$$

$$x=\frac{7-\sqrt{33}}{4}\approx 0.313859338$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+1.

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

Use the distributive property to multiply 3 by 3x+1.

$$2x^{2}+2x+5=9x+3$$

Subtract 9x from both sides.

$$2x^{2}+2x+5-9x=3$$

Combine 2x and -9x to get -7x.

$$2x^{2}-7x+5=3$$

Subtract 3 from both sides.

$$2x^{2}-7x+5-3=0$$

Subtract 3 from 5 to get 2.

$$2x^{2}-7x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-16}}{2\times 2}$$

Add 49 to -16.

$$x=\frac{-\left(-7\right)±\sqrt{33}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{33}}{4}$$

Now solve the equation x=\frac{7±\sqrt{33}}{4} when ± is plus. Add 7 to \sqrt{33}.

$$x=\frac{\sqrt{33}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{33}}{4} when ± is minus. Subtract \sqrt{33} from 7.

$$x=\frac{7-\sqrt{33}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+7}{4} x=\frac{7-\sqrt{33}}{4}$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+1.

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

Use the distributive property to multiply 3 by 3x+1.

$$2x^{2}+2x+5=9x+3$$

Subtract 9x from both sides.

$$2x^{2}+2x+5-9x=3$$

Combine 2x and -9x to get -7x.

$$2x^{2}-7x+5=3$$

Subtract 5 from both sides.

$$2x^{2}-7x=3-5$$

Subtract 5 from 3 to get -2.

$$2x^{2}-7x=-2$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=-\frac{2}{2}$$

Divide -2 by 2.

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

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-1+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-1+\frac{49}{16}$$

Add -1 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{33}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{33}}{4} x-\frac{7}{4}=-\frac{\sqrt{33}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{\sqrt{33}+7}{4} x=\frac{7-\sqrt{33}}{4}$$