Solve 3/2x+3=x+2

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

Solve for x

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

$$x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+3.

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

Use the distributive property to multiply 2x+3 by x.

$$3=2x^{2}+3x+\left(2x+3\right)\times 2$$

Use the distributive property to multiply 2x+3 by 2.

$$3=2x^{2}+3x+4x+6$$

Combine 3x and 4x to get 7x.

$$3=2x^{2}+7x+6$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+7x+6=3$$

Subtract 3 from both sides.

$$2x^{2}+7x+6-3=0$$

Subtract 3 from 6 to get 3.

$$2x^{2}+7x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 3}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$x=\frac{-7±\sqrt{49-24}}{2\times 2}$$

Add 49 to -24.

$$x=\frac{-7±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-7±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±5}{4}$$

Now solve the equation x=\frac{-7±5}{4} when ± is plus. Add -7 to 5.

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

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

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

Now solve the equation x=\frac{-7±5}{4} when ± is minus. Subtract 5 from -7.

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

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to -\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+3.

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

Use the distributive property to multiply 2x+3 by x.

$$3=2x^{2}+3x+\left(2x+3\right)\times 2$$

Use the distributive property to multiply 2x+3 by 2.

$$3=2x^{2}+3x+4x+6$$

Combine 3x and 4x to get 7x.

$$3=2x^{2}+7x+6$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+7x+6=3$$

Subtract 6 from both sides.

$$2x^{2}+7x=3-6$$

Subtract 6 from 3 to get -3.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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.

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

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

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

Add -\frac{3}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{25}{16}$$

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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{5}{4} x+\frac{7}{4}=-\frac{5}{4}$$

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

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