Solve x^2+3x-3+5=0

$${ x }^{ 2 } +3x-3+5=0$$

Solve for x

$$x=-2$$

$$x=-1$$

Steps Using Factoring

Add -3 and 5 to get 2.

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

To solve the equation, factor x^{2}+3x+2 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

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

$$x=-1 x=-2$$

Steps Using Factoring By Grouping

Add -3 and 5 to get 2.

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

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

$$a+b=3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

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

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

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

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

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

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

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

$$x=-1 x=-2$$

Steps Using the Quadratic Formula

Add -3 and 5 to get 2.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8}}{2}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-3±1}{2}$$

Now solve the equation x=\frac{-3±1}{2} when ± is plus. Add -3 to 1.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-3±1}{2} when ± is minus. Subtract 1 from -3.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=-1 x=-2$$

Steps for Completing the Square

Add -3 and 5 to get 2.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

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

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

Add -2 to \frac{9}{4}.

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

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.

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

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

$$x=-1 x=-2$$