Solve x-3(x+2)^2=-3*2

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

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

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

Use the distributive property to multiply -3 by x^{2}+4x+4.

$$x-3x^{2}-12x-12=-3\times 2$$

Combine x and -12x to get -11x.

$$-11x-3x^{2}-12=-3\times 2$$

Multiply -3 and 2 to get -6.

$$-11x-3x^{2}-12=-6$$

Add 6 to both sides.

$$-11x-3x^{2}-12+6=0$$

Add -12 and 6 to get -6.

$$-11x-3x^{2}-6=0$$

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

$$-3x^{2}-11x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=-3\left(-6\right)=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -11.

$$a=-2 b=-9$$

Rewrite -3x^{2}-11x-6 as \left(-3x^{2}-2x\right)+\left(-9x-6\right).

$$\left(-3x^{2}-2x\right)+\left(-9x-6\right)$$

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

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

Factor out common term 3x+2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

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

Use the distributive property to multiply -3 by x^{2}+4x+4.

$$x-3x^{2}-12x-12=-3\times 2$$

Combine x and -12x to get -11x.

$$-11x-3x^{2}-12=-3\times 2$$

Multiply -3 and 2 to get -6.

$$-11x-3x^{2}-12=-6$$

Add 6 to both sides.

$$-11x-3x^{2}-12+6=0$$

Add -12 and 6 to get -6.

$$-11x-3x^{2}-6=0$$

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.

$$-3x^{2}-11x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -11 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-3\right)\left(-6\right)}}{2\left(-3\right)}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-3\right)\left(-6\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-11\right)±\sqrt{121+12\left(-6\right)}}{2\left(-3\right)}$$

Multiply 12 times -6.

$$x=\frac{-\left(-11\right)±\sqrt{121-72}}{2\left(-3\right)}$$

Add 121 to -72.

$$x=\frac{-\left(-11\right)±\sqrt{49}}{2\left(-3\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-11\right)±7}{2\left(-3\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±7}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{11±7}{-6}$$

Now solve the equation x=\frac{11±7}{-6} when ± is plus. Add 11 to 7.

$$x=\frac{18}{-6}$$

Divide 18 by -6.

$$x=-3$$

Now solve the equation x=\frac{11±7}{-6} when ± is minus. Subtract 7 from 11.

$$x=\frac{4}{-6}$$

Reduce the fraction \frac{4}{-6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

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

Use the distributive property to multiply -3 by x^{2}+4x+4.

$$x-3x^{2}-12x-12=-3\times 2$$

Combine x and -12x to get -11x.

$$-11x-3x^{2}-12=-3\times 2$$

Multiply -3 and 2 to get -6.

$$-11x-3x^{2}-12=-6$$

Add 12 to both sides.

$$-11x-3x^{2}=-6+12$$

Add -6 and 12 to get 6.

$$-11x-3x^{2}=6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-3x^{2}-11x=6$$

Divide both sides by -3.

$$\frac{-3x^{2}-11x}{-3}=\frac{6}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{11}{-3}\right)x=\frac{6}{-3}$$

Divide -11 by -3.

$$x^{2}+\frac{11}{3}x=\frac{6}{-3}$$

Divide 6 by -3.

$$x^{2}+\frac{11}{3}x=-2$$

Divide \frac{11}{3}, the coefficient of the x term, by 2 to get \frac{11}{6}. Then add the square of \frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{3}x+\left(\frac{11}{6}\right)^{2}=-2+\left(\frac{11}{6}\right)^{2}$$

Square \frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=-2+\frac{121}{36}$$

Add -2 to \frac{121}{36}.

$$x^{2}+\frac{11}{3}x+\frac{121}{36}=\frac{49}{36}$$

Factor x^{2}+\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x+\frac{11}{6}=\frac{7}{6} x+\frac{11}{6}=-\frac{7}{6}$$

Subtract \frac{11}{6} from both sides of the equation.

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