Solve -x^2-3x/x+3=1/2

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

Solve for x

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

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 2\left(x+3\right), the least common multiple of x+3,2.

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

Use the distributive property to multiply 2 by -x^{2}-3x.

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

Subtract x from both sides.

$$2\left(-x^{2}\right)-6x-x=3$$

Combine -6x and -x to get -7x.

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

Subtract 3 from both sides.

$$2\left(-x^{2}\right)-7x-3=0$$

Multiply 2 and -1 to get -2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-2\left(-3\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-1 b=-6$$

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

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

Factor out -x in the first and -3 in the second group.

$$-x\left(2x+1\right)-3\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(-x-3\right)$$

To find equation solutions, solve 2x+1=0 and -x-3=0.

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

Variable x cannot be equal to -3.

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

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 2\left(x+3\right), the least common multiple of x+3,2.

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

Use the distributive property to multiply 2 by -x^{2}-3x.

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

Subtract x from both sides.

$$2\left(-x^{2}\right)-6x-x=3$$

Combine -6x and -x to get -7x.

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

Subtract 3 from both sides.

$$2\left(-x^{2}\right)-7x-3=0$$

Multiply 2 and -1 to get -2.

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

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}.

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

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-7\right)±\sqrt{49+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$x=\frac{-\left(-7\right)±\sqrt{49-24}}{2\left(-2\right)}$$

Add 49 to -24.

$$x=\frac{-\left(-7\right)±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-7\right)±5}{2\left(-2\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±5}{2\left(-2\right)}$$

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.

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

Divide 12 by -4.

$$x=-3$$

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

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

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

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

The equation is now solved.

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

Variable x cannot be equal to -3.

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

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 2\left(x+3\right), the least common multiple of x+3,2.

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

Use the distributive property to multiply 2 by -x^{2}-3x.

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

Subtract x from both sides.

$$2\left(-x^{2}\right)-6x-x=3$$

Combine -6x and -x to get -7x.

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

Multiply 2 and -1 to get -2.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -7 by -2.

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

Divide 3 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.

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

Variable x cannot be equal to -3.

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