Solve 6x^2+3x-50-x=2x^2+106

$$6 x ^ { 2 } + 3 x - 50 - x = 2 x ^ { 2 } + 106$$

Solve for x

$$x = -\frac{13}{2} = -6\frac{1}{2} = -6.5$$

$$x=6$$

Steps Using Factoring By Grouping

Combine 3x and -x to get 2x.

$$6x^{2}+2x-50=2x^{2}+106$$

Subtract 2x^{2} from both sides.

$$6x^{2}+2x-50-2x^{2}=106$$

Combine 6x^{2} and -2x^{2} to get 4x^{2}.

$$4x^{2}+2x-50=106$$

Subtract 106 from both sides.

$$4x^{2}+2x-50-106=0$$

Subtract 106 from -50 to get -156.

$$4x^{2}+2x-156=0$$

Divide both sides by 2.

$$2x^{2}+x-78=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-78. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-78\right)=-156$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -156.

$$-1,156 -2,78 -3,52 -4,39 -6,26 -12,13$$

Calculate the sum for each pair.

$$-1+156=155 -2+78=76 -3+52=49 -4+39=35 -6+26=20 -12+13=1$$

The solution is the pair that gives sum 1.

$$a=-12 b=13$$

Rewrite 2x^{2}+x-78 as \left(2x^{2}-12x\right)+\left(13x-78\right).

$$\left(2x^{2}-12x\right)+\left(13x-78\right)$$

Factor out 2x in the first and 13 in the second group.

$$2x\left(x-6\right)+13\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(2x+13\right)$$

To find equation solutions, solve x-6=0 and 2x+13=0.

$$x=6 x=-\frac{13}{2}$$

Steps Using the Quadratic Formula

Combine 3x and -x to get 2x.

$$6x^{2}+2x-50=2x^{2}+106$$

Subtract 2x^{2} from both sides.

$$6x^{2}+2x-50-2x^{2}=106$$

Combine 6x^{2} and -2x^{2} to get 4x^{2}.

$$4x^{2}+2x-50=106$$

Subtract 106 from both sides.

$$4x^{2}+2x-50-106=0$$

Subtract 106 from -50 to get -156.

$$4x^{2}+2x-156=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 2 for b, and -156 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\left(-156\right)}}{2\times 4}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 4\left(-156\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-2±\sqrt{4-16\left(-156\right)}}{2\times 4}$$

Multiply -16 times -156.

$$x=\frac{-2±\sqrt{4+2496}}{2\times 4}$$

Add 4 to 2496.

$$x=\frac{-2±\sqrt{2500}}{2\times 4}$$

Take the square root of 2500.

$$x=\frac{-2±50}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±50}{8}$$

Now solve the equation x=\frac{-2±50}{8} when ± is plus. Add -2 to 50.

$$x=\frac{48}{8}$$

Divide 48 by 8.

$$x=6$$

Now solve the equation x=\frac{-2±50}{8} when ± is minus. Subtract 50 from -2.

$$x=-\frac{52}{8}$$

Reduce the fraction \frac{-52}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{13}{2}$$

The equation is now solved.

$$x=6 x=-\frac{13}{2}$$

Steps for Completing the Square

Combine 3x and -x to get 2x.

$$6x^{2}+2x-50=2x^{2}+106$$

Subtract 2x^{2} from both sides.

$$6x^{2}+2x-50-2x^{2}=106$$

Combine 6x^{2} and -2x^{2} to get 4x^{2}.

$$4x^{2}+2x-50=106$$

Add 50 to both sides.

$$4x^{2}+2x=106+50$$

Add 106 and 50 to get 156.

$$4x^{2}+2x=156$$

Divide both sides by 4.

$$\frac{4x^{2}+2x}{4}=\frac{156}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{2}{4}x=\frac{156}{4}$$

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

$$x^{2}+\frac{1}{2}x=\frac{156}{4}$$

Divide 156 by 4.

$$x^{2}+\frac{1}{2}x=39$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=39+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=39+\frac{1}{16}$$

Add 39 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{625}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{625}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{625}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{25}{4} x+\frac{1}{4}=-\frac{25}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=6 x=-\frac{13}{2}$$