Solve 2x^2+x-15/x+2=0

$$\frac { 2 x ^ { 2 } + x - 15 } { x + 2 } = 0$$

Solve for x

$$x=-3$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-15\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite 2x^{2}+x-15 as \left(2x^{2}-5x\right)+\left(6x-15\right).

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

Factor out x in the first and 3 in the second group.

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

Factor out common term 2x-5 by using distributive property.

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

To find equation solutions, solve 2x-5=0 and x+3=0.

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

Steps Using the Quadratic Formula

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-15\right)}}{2\times 2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-15\right)}}{2\times 2}$$

Multiply -8 times -15.

$$x=\frac{-1±\sqrt{1+120}}{2\times 2}$$

Add 1 to 120.

$$x=\frac{-1±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-1±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±11}{4}$$

Now solve the equation x=\frac{-1±11}{4} when ± is plus. Add -1 to 11.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{-1±11}{4} when ± is minus. Subtract 11 from -1.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

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

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

Add 15 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{1}{2}x=\frac{15}{2}$$

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.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{15}{2}+\left(\frac{1}{4}\right)^{2}$$

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{15}{2}+\frac{1}{16}$$

Add \frac{15}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{121}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

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

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

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