Solve 2x^2-9x-5/x+3=0

$$\frac { 2 x ^ { 2 } - 9 x - 5 } { x + 3 } = 0$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=5$$

Steps Using Factoring By Grouping

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

$$2x^{2}-9x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-9 ab=2\left(-5\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$a=-10 b=1$$

Rewrite 2x^{2}-9x-5 as \left(2x^{2}-10x\right)+\left(x-5\right).

$$\left(2x^{2}-10x\right)+\left(x-5\right)$$

Factor out 2x in 2x^{2}-10x.

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

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

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

To find equation solutions, solve x-5=0 and 2x+1=0.

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

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.

$$2x^{2}-9x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-\left(-9\right)±\sqrt{81+40}}{2\times 2}$$

Add 81 to 40.

$$x=\frac{-\left(-9\right)±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-\left(-9\right)±11}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±11}{4}$$

Now solve the equation x=\frac{9±11}{4} when ± is plus. Add 9 to 11.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{9±11}{4} when ± is minus. Subtract 11 from 9.

$$x=-\frac{2}{4}$$

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

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

The equation is now solved.

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

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.

$$2x^{2}-9x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

$$\frac{2x^{2}-9x}{2}=\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{9}{2}x=\frac{5}{2}$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=\frac{5}{2}+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{5}{2}+\frac{81}{16}$$

Add \frac{5}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{121}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{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}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{11}{4} x-\frac{9}{4}=-\frac{11}{4}$$

Add \frac{9}{4} to both sides of the equation.

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