Solve x-6+2x^2=2-x^2+3x

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

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 2 from both sides.

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

Subtract 2 from -6 to get -8.

$$x-8+2x^{2}=-x^{2}+3x$$

Add x^{2} to both sides.

$$x-8+2x^{2}+x^{2}=3x$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$x-8+3x^{2}=3x$$

Subtract 3x from both sides.

$$x-8+3x^{2}-3x=0$$

Combine x and -3x to get -2x.

$$-2x-8+3x^{2}=0$$

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

$$3x^{2}-2x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-8\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -2.

$$a=-6 b=4$$

Rewrite 3x^{2}-2x-8 as \left(3x^{2}-6x\right)+\left(4x-8\right).

$$\left(3x^{2}-6x\right)+\left(4x-8\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(x-2\right)+4\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(3x+4\right)$$

To find equation solutions, solve x-2=0 and 3x+4=0.

$$x=2 x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

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

Subtract 2 from -6 to get -8.

$$x-8+2x^{2}=-x^{2}+3x$$

Add x^{2} to both sides.

$$x-8+2x^{2}+x^{2}=3x$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$x-8+3x^{2}=3x$$

Subtract 3x from both sides.

$$x-8+3x^{2}-3x=0$$

Combine x and -3x to get -2x.

$$-2x-8+3x^{2}=0$$

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.

$$3x^{2}-2x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-8\right)}}{2\times 3}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\times 3}$$

Add 4 to 96.

$$x=\frac{-\left(-2\right)±\sqrt{100}}{2\times 3}$$

Take the square root of 100.

$$x=\frac{-\left(-2\right)±10}{2\times 3}$$

The opposite of -2 is 2.

$$x=\frac{2±10}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±10}{6}$$

Now solve the equation x=\frac{2±10}{6} when ± is plus. Add 2 to 10.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{2±10}{6} when ± is minus. Subtract 10 from 2.

$$x=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{3}$$

The equation is now solved.

$$x=2 x=-\frac{4}{3}$$

Steps for Completing the Square

Add x^{2} to both sides.

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

Combine 2x^{2} and x^{2} to get 3x^{2}.

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

Subtract 3x from both sides.

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

Combine x and -3x to get -2x.

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

Add 6 to both sides.

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

Add 2 and 6 to get 8.

$$-2x+3x^{2}=8$$

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.

$$3x^{2}-2x=8$$

Divide both sides by 3.

$$\frac{3x^{2}-2x}{3}=\frac{8}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{2}{3}x=\frac{8}{3}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{8}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{8}{3}+\frac{1}{9}$$

Add \frac{8}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{25}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{5}{3} x-\frac{1}{3}=-\frac{5}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=2 x=-\frac{4}{3}$$