Solve 14x^2-(2x-11)(5x+4)-33x=32

$$14 { x }^{ 2 } -(2x-11)(5x+4)-33x=32$$

Solve for x

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

$$x=-2$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x-11 by 5x+4 and combine like terms.

$$14x^{2}-\left(10x^{2}-47x-44\right)-33x=32$$

To find the opposite of 10x^{2}-47x-44, find the opposite of each term.

$$14x^{2}-10x^{2}+47x+44-33x=32$$

Combine 14x^{2} and -10x^{2} to get 4x^{2}.

$$4x^{2}+47x+44-33x=32$$

Combine 47x and -33x to get 14x.

$$4x^{2}+14x+44=32$$

Subtract 32 from both sides.

$$4x^{2}+14x+44-32=0$$

Subtract 32 from 44 to get 12.

$$4x^{2}+14x+12=0$$

Divide both sides by 2.

$$2x^{2}+7x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\times 6=12$$

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 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 2x^{2}+7x+6 as \left(2x^{2}+3x\right)+\left(4x+6\right).

$$\left(2x^{2}+3x\right)+\left(4x+6\right)$$

Factor out x in the first and 2 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-11 by 5x+4 and combine like terms.

$$14x^{2}-\left(10x^{2}-47x-44\right)-33x=32$$

To find the opposite of 10x^{2}-47x-44, find the opposite of each term.

$$14x^{2}-10x^{2}+47x+44-33x=32$$

Combine 14x^{2} and -10x^{2} to get 4x^{2}.

$$4x^{2}+47x+44-33x=32$$

Combine 47x and -33x to get 14x.

$$4x^{2}+14x+44=32$$

Subtract 32 from both sides.

$$4x^{2}+14x+44-32=0$$

Subtract 32 from 44 to get 12.

$$4x^{2}+14x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 14 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\times 4\times 12}}{2\times 4}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 4\times 12}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-14±\sqrt{196-16\times 12}}{2\times 4}$$

Multiply -16 times 12.

$$x=\frac{-14±\sqrt{196-192}}{2\times 4}$$

Add 196 to -192.

$$x=\frac{-14±\sqrt{4}}{2\times 4}$$

Take the square root of 4.

$$x=\frac{-14±2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-14±2}{8}$$

Now solve the equation x=\frac{-14±2}{8} when ± is plus. Add -14 to 2.

$$x=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-14±2}{8} when ± is minus. Subtract 2 from -14.

$$x=-\frac{16}{8}$$

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 2x-11 by 5x+4 and combine like terms.

$$14x^{2}-\left(10x^{2}-47x-44\right)-33x=32$$

To find the opposite of 10x^{2}-47x-44, find the opposite of each term.

$$14x^{2}-10x^{2}+47x+44-33x=32$$

Combine 14x^{2} and -10x^{2} to get 4x^{2}.

$$4x^{2}+47x+44-33x=32$$

Combine 47x and -33x to get 14x.

$$4x^{2}+14x+44=32$$

Subtract 44 from both sides.

$$4x^{2}+14x=32-44$$

Subtract 44 from 32 to get -12.

$$4x^{2}+14x=-12$$

Divide both sides by 4.

$$\frac{4x^{2}+14x}{4}=-\frac{12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{14}{4}x=-\frac{12}{4}$$

Reduce the fraction \frac{14}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{7}{2}x=-\frac{12}{4}$$

Divide -12 by 4.

$$x^{2}+\frac{7}{2}x=-3$$

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.

$$x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=-3+\left(\frac{7}{4}\right)^{2}$$

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

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=-3+\frac{49}{16}$$

Add -3 to \frac{49}{16}.

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

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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{1}{4} x+\frac{7}{4}=-\frac{1}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

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