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

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

Solve for x

$$x=\frac{\sqrt{13}-3}{2}\approx 0.302775638$$

$$x=\frac{-\sqrt{13}-3}{2}\approx -3.302775638$$

Steps Using the Quadratic Formula

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

$$2x+1=-5x\left(x+3\right)+\left(x+3\right)\times 2$$

Use the distributive property to multiply -5x by x+3.

$$2x+1=-5x^{2}-15x+\left(x+3\right)\times 2$$

Use the distributive property to multiply x+3 by 2.

$$2x+1=-5x^{2}-15x+2x+6$$

Combine -15x and 2x to get -13x.

$$2x+1=-5x^{2}-13x+6$$

Add 5x^{2} to both sides.

$$2x+1+5x^{2}=-13x+6$$

Add 13x to both sides.

$$2x+1+5x^{2}+13x=6$$

Combine 2x and 13x to get 15x.

$$15x+1+5x^{2}=6$$

Subtract 6 from both sides.

$$15x+1+5x^{2}-6=0$$

Subtract 6 from 1 to get -5.

$$15x-5+5x^{2}=0$$

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.

$$5x^{2}+15x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 15 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\times 5\left(-5\right)}}{2\times 5}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 5\left(-5\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-15±\sqrt{225-20\left(-5\right)}}{2\times 5}$$

Multiply -20 times -5.

$$x=\frac{-15±\sqrt{225+100}}{2\times 5}$$

Add 225 to 100.

$$x=\frac{-15±\sqrt{325}}{2\times 5}$$

Take the square root of 325.

$$x=\frac{-15±5\sqrt{13}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-15±5\sqrt{13}}{10}$$

Now solve the equation x=\frac{-15±5\sqrt{13}}{10} when ± is plus. Add -15 to 5\sqrt{13}.

$$x=\frac{5\sqrt{13}-15}{10}$$

Divide -15+5\sqrt{13} by 10.

$$x=\frac{\sqrt{13}-3}{2}$$

Now solve the equation x=\frac{-15±5\sqrt{13}}{10} when ± is minus. Subtract 5\sqrt{13} from -15.

$$x=\frac{-5\sqrt{13}-15}{10}$$

Divide -15-5\sqrt{13} by 10.

$$x=\frac{-\sqrt{13}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{13}-3}{2} x=\frac{-\sqrt{13}-3}{2}$$

Steps for Completing the Square

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

$$2x+1=-5x\left(x+3\right)+\left(x+3\right)\times 2$$

Use the distributive property to multiply -5x by x+3.

$$2x+1=-5x^{2}-15x+\left(x+3\right)\times 2$$

Use the distributive property to multiply x+3 by 2.

$$2x+1=-5x^{2}-15x+2x+6$$

Combine -15x and 2x to get -13x.

$$2x+1=-5x^{2}-13x+6$$

Add 5x^{2} to both sides.

$$2x+1+5x^{2}=-13x+6$$

Add 13x to both sides.

$$2x+1+5x^{2}+13x=6$$

Combine 2x and 13x to get 15x.

$$15x+1+5x^{2}=6$$

Subtract 1 from both sides.

$$15x+5x^{2}=6-1$$

Subtract 1 from 6 to get 5.

$$15x+5x^{2}=5$$

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.

$$5x^{2}+15x=5$$

Divide both sides by 5.

$$\frac{5x^{2}+15x}{5}=\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{15}{5}x=\frac{5}{5}$$

Divide 15 by 5.

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

Divide 5 by 5.

$$x^{2}+3x=1$$

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=1+\frac{9}{4}$$

Add 1 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{13}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{13}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{13}}{2} x+\frac{3}{2}=-\frac{\sqrt{13}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{13}-3}{2} x=\frac{-\sqrt{13}-3}{2}$$