Solve x^2+11x/2+6=0

$${ x }^{ 2 } + \frac{ 11x }{ 2 } +6=0$$

Solve for x

$$x=-4$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

$$2x^{2}+11x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=11 ab=2\times 12=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=3 b=8$$

Rewrite 2x^{2}+11x+12 as \left(2x^{2}+3x\right)+\left(8x+12\right).

$$\left(2x^{2}+3x\right)+\left(8x+12\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(2x+3\right)+4\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(x+4\right)$$

To find equation solutions, solve 2x+3=0 and x+4=0.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2x^{2}+11x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 11 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 2\times 12}}{2\times 2}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 2\times 12}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-11±\sqrt{121-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$x=\frac{-11±\sqrt{121-96}}{2\times 2}$$

Add 121 to -96.

$$x=\frac{-11±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-11±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-11±5}{4}$$

Now solve the equation x=\frac{-11±5}{4} when ± is plus. Add -11 to 5.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-11±5}{4} when ± is minus. Subtract 5 from -11.

$$x=-\frac{16}{4}$$

Divide -16 by 4.

$$x=-4$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$2x^{2}+11x+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+11x=-12$$

Divide both sides by 2.

$$\frac{2x^{2}+11x}{2}=-\frac{12}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{11}{2}x=-\frac{12}{2}$$

Divide -12 by 2.

$$x^{2}+\frac{11}{2}x=-6$$

Divide \frac{11}{2}, the coefficient of the x term, by 2 to get \frac{11}{4}. Then add the square of \frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{2}x+\left(\frac{11}{4}\right)^{2}=-6+\left(\frac{11}{4}\right)^{2}$$

Square \frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=-6+\frac{121}{16}$$

Add -6 to \frac{121}{16}.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{25}{16}$$

Factor x^{2}+\frac{11}{2}x+\frac{121}{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}.

$$\left(x+\frac{11}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{11}{4}=\frac{5}{4} x+\frac{11}{4}=-\frac{5}{4}$$

Subtract \frac{11}{4} from both sides of the equation.

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