Solve (3x+5/2)(2x^2+6x+4)-(4x+6)(3/2x^2+5/2x)/(2x^2+6x+4)^2=0

$$\frac { ( 3 x + \frac { 5 } { 2 } ) ( 2 x ^ { 2 } + 6 x + 4 ) - ( 4 x + 6 ) ( \frac { 3 } { 2 } x ^ { 2 } + \frac { 5 } { 2 } x ) } { ( 2 x ^ { 2 } + 6 x + 4 ) ^ { 2 } } = 0$$

Solve for x (complex solution)

$$x=-\frac{3}{2}+\frac{1}{2}i=-1.5+0.5i$$

$$x=-\frac{3}{2}-\frac{1}{2}i=-1.5-0.5i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,-1 since division by zero is not defined. Multiply both sides of the equation by 4\left(x+1\right)^{2}\left(x+2\right)^{2}.

$$\left(3x+\frac{5}{2}\right)\left(2x^{2}+6x+4\right)-\left(4x+6\right)\left(\frac{3}{2}x^{2}+\frac{5}{2}x\right)=0$$

Use the distributive property to multiply 3x+\frac{5}{2} by 2x^{2}+6x+4 and combine like terms.

$$6x^{3}+23x^{2}+27x+10-\left(4x+6\right)\left(\frac{3}{2}x^{2}+\frac{5}{2}x\right)=0$$

Use the distributive property to multiply 4x+6 by \frac{3}{2}x^{2}+\frac{5}{2}x and combine like terms.

$$6x^{3}+23x^{2}+27x+10-\left(6x^{3}+19x^{2}+15x\right)=0$$

To find the opposite of 6x^{3}+19x^{2}+15x, find the opposite of each term.

$$6x^{3}+23x^{2}+27x+10-6x^{3}-19x^{2}-15x=0$$

Combine 6x^{3} and -6x^{3} to get 0.

$$23x^{2}+27x+10-19x^{2}-15x=0$$

Combine 23x^{2} and -19x^{2} to get 4x^{2}.

$$4x^{2}+27x+10-15x=0$$

Combine 27x and -15x to get 12x.

$$4x^{2}+12x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 12 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\times 10}}{2\times 4}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 4\times 10}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-12±\sqrt{144-16\times 10}}{2\times 4}$$

Multiply -16 times 10.

$$x=\frac{-12±\sqrt{144-160}}{2\times 4}$$

Add 144 to -160.

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

Take the square root of -16.

$$x=\frac{-12±4i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-12±4i}{8}$$

Now solve the equation x=\frac{-12±4i}{8} when ± is plus. Add -12 to 4i.

$$x=\frac{-12+4i}{8}$$

Divide -12+4i by 8.

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

Now solve the equation x=\frac{-12±4i}{8} when ± is minus. Subtract 4i from -12.

$$x=\frac{-12-4i}{8}$$

Divide -12-4i by 8.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,-1 since division by zero is not defined. Multiply both sides of the equation by 4\left(x+1\right)^{2}\left(x+2\right)^{2}.

$$\left(3x+\frac{5}{2}\right)\left(2x^{2}+6x+4\right)-\left(4x+6\right)\left(\frac{3}{2}x^{2}+\frac{5}{2}x\right)=0$$

Use the distributive property to multiply 3x+\frac{5}{2} by 2x^{2}+6x+4 and combine like terms.

$$6x^{3}+23x^{2}+27x+10-\left(4x+6\right)\left(\frac{3}{2}x^{2}+\frac{5}{2}x\right)=0$$

Use the distributive property to multiply 4x+6 by \frac{3}{2}x^{2}+\frac{5}{2}x and combine like terms.

$$6x^{3}+23x^{2}+27x+10-\left(6x^{3}+19x^{2}+15x\right)=0$$

To find the opposite of 6x^{3}+19x^{2}+15x, find the opposite of each term.

$$6x^{3}+23x^{2}+27x+10-6x^{3}-19x^{2}-15x=0$$

Combine 6x^{3} and -6x^{3} to get 0.

$$23x^{2}+27x+10-19x^{2}-15x=0$$

Combine 23x^{2} and -19x^{2} to get 4x^{2}.

$$4x^{2}+27x+10-15x=0$$

Combine 27x and -15x to get 12x.

$$4x^{2}+12x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+12x=-10$$

Divide both sides by 4.

$$\frac{4x^{2}+12x}{4}=-\frac{10}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{12}{4}x=-\frac{10}{4}$$

Divide 12 by 4.

$$x^{2}+3x=-\frac{10}{4}$$

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

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

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.

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

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

$$x^{2}+3x+\frac{9}{4}=-\frac{5}{2}+\frac{9}{4}$$

Add -\frac{5}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{1}{4}}$$

Simplify.

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

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

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