Solve 3x^2+3=-6x

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

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

Add 6x to both sides.

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

Divide both sides by 3.

$$x^{2}+1+2x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

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

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.

$$a=1 b=1$$

Rewrite x^{2}+2x+1 as \left(x^{2}+x\right)+\left(x+1\right).

$$\left(x^{2}+x\right)+\left(x+1\right)$$

Factor out x in x^{2}+x.

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

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

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

Rewrite as a binomial square.

$$\left(x+1\right)^{2}$$

To find equation solution, solve x+1=0.

$$x=-1$$

Steps Using the Quadratic Formula

Add 6x to both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 3\times 3}}{2\times 3}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-6±\sqrt{36-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$x=\frac{-6±\sqrt{36-36}}{2\times 3}$$

Add 36 to -36.

$$x=\frac{-6±\sqrt{0}}{2\times 3}$$

Take the square root of 0.

$$x=-\frac{6}{2\times 3}$$

Multiply 2 times 3.

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

Steps for Completing the Square

Add 6x to both sides.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 3.

$$\frac{3x^{2}+6x}{3}=-\frac{3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{6}{3}x=-\frac{3}{3}$$

Divide 6 by 3.

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

Divide -3 by 3.

$$x^{2}+2x=-1$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=-1+1^{2}$$

Square 1.

$$x^{2}+2x+1=-1+1$$

Add -1 to 1.

$$x^{2}+2x+1=0$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+1=0 x+1=0$$

Subtract 1 from both sides of the equation.

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

The equation is now solved. Solutions are the same.

$$x=-1$$