Solve 2x^2+10x+6=-7x^2-14x-10

$$2 { x }^{ 2 } +10x+6 = -7 { x }^{ 2 } -14x-10$$

Solve for x

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

Steps Using Factoring By Grouping

Add 7x^{2} to both sides.

$$2x^{2}+10x+6+7x^{2}=-14x-10$$

Combine 2x^{2} and 7x^{2} to get 9x^{2}.

$$9x^{2}+10x+6=-14x-10$$

Add 14x to both sides.

$$9x^{2}+10x+6+14x=-10$$

Combine 10x and 14x to get 24x.

$$9x^{2}+24x+6=-10$$

Add 10 to both sides.

$$9x^{2}+24x+6+10=0$$

Add 6 and 10 to get 16.

$$9x^{2}+24x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=24 ab=9\times 16=144$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 144.

$$1,144 2,72 3,48 4,36 6,24 8,18 9,16 12,12$$

Calculate the sum for each pair.

$$1+144=145 2+72=74 3+48=51 4+36=40 6+24=30 8+18=26 9+16=25 12+12=24$$

The solution is the pair that gives sum 24.

$$a=12 b=12$$

Rewrite 9x^{2}+24x+16 as \left(9x^{2}+12x\right)+\left(12x+16\right).

$$\left(9x^{2}+12x\right)+\left(12x+16\right)$$

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

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

Factor out common term 3x+4 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve 3x+4=0.

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

Steps Using the Quadratic Formula

Add 7x^{2} to both sides.

$$2x^{2}+10x+6+7x^{2}=-14x-10$$

Combine 2x^{2} and 7x^{2} to get 9x^{2}.

$$9x^{2}+10x+6=-14x-10$$

Add 14x to both sides.

$$9x^{2}+10x+6+14x=-10$$

Combine 10x and 14x to get 24x.

$$9x^{2}+24x+6=-10$$

Add 10 to both sides.

$$9x^{2}+24x+6+10=0$$

Add 6 and 10 to get 16.

$$9x^{2}+24x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 24 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\times 9\times 16}}{2\times 9}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 9\times 16}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-24±\sqrt{576-36\times 16}}{2\times 9}$$

Multiply -36 times 16.

$$x=\frac{-24±\sqrt{576-576}}{2\times 9}$$

Add 576 to -576.

$$x=\frac{-24±\sqrt{0}}{2\times 9}$$

Take the square root of 0.

$$x=-\frac{24}{2\times 9}$$

Multiply 2 times 9.

$$x=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

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

Steps for Completing the Square

Add 7x^{2} to both sides.

$$2x^{2}+10x+6+7x^{2}=-14x-10$$

Combine 2x^{2} and 7x^{2} to get 9x^{2}.

$$9x^{2}+10x+6=-14x-10$$

Add 14x to both sides.

$$9x^{2}+10x+6+14x=-10$$

Combine 10x and 14x to get 24x.

$$9x^{2}+24x+6=-10$$

Subtract 6 from both sides.

$$9x^{2}+24x=-10-6$$

Subtract 6 from -10 to get -16.

$$9x^{2}+24x=-16$$

Divide both sides by 9.

$$\frac{9x^{2}+24x}{9}=-\frac{16}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{24}{9}x=-\frac{16}{9}$$

Reduce the fraction \frac{24}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{8}{3}x=-\frac{16}{9}$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=-\frac{16}{9}+\left(\frac{4}{3}\right)^{2}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{-16+16}{9}$$

Add -\frac{16}{9} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=0$$

Factor x^{2}+\frac{8}{3}x+\frac{16}{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}.

$$\left(x+\frac{4}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\frac{4}{3}=0 x+\frac{4}{3}=0$$

Subtract \frac{4}{3} from both sides of the equation.

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

The equation is now solved. Solutions are the same.

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