Solve 10x^2+10x+8=3x^2-10x+11

$$10 { x }^{ 2 } +10x+8=3 { x }^{ 2 } -10x+11$$

Solve for x

$$x=-3$$

$$x=\frac{1}{7}\approx 0.142857143$$

Steps Using Factoring By Grouping

Subtract 3x^{2} from both sides.

$$10x^{2}+10x+8-3x^{2}=-10x+11$$

Combine 10x^{2} and -3x^{2} to get 7x^{2}.

$$7x^{2}+10x+8=-10x+11$$

Add 10x to both sides.

$$7x^{2}+10x+8+10x=11$$

Combine 10x and 10x to get 20x.

$$7x^{2}+20x+8=11$$

Subtract 11 from both sides.

$$7x^{2}+20x+8-11=0$$

Subtract 11 from 8 to get -3.

$$7x^{2}+20x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=20 ab=7\left(-3\right)=-21$$

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 -21.

$$-1,21 -3,7$$

Calculate the sum for each pair.

$$-1+21=20 -3+7=4$$

The solution is the pair that gives sum 20.

$$a=-1 b=21$$

Rewrite 7x^{2}+20x-3 as \left(7x^{2}-x\right)+\left(21x-3\right).

$$\left(7x^{2}-x\right)+\left(21x-3\right)$$

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

$$x\left(7x-1\right)+3\left(7x-1\right)$$

Factor out common term 7x-1 by using distributive property.

$$\left(7x-1\right)\left(x+3\right)$$

To find equation solutions, solve 7x-1=0 and x+3=0.

$$x=\frac{1}{7} x=-3$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$10x^{2}+10x+8-3x^{2}=-10x+11$$

Combine 10x^{2} and -3x^{2} to get 7x^{2}.

$$7x^{2}+10x+8=-10x+11$$

Add 10x to both sides.

$$7x^{2}+10x+8+10x=11$$

Combine 10x and 10x to get 20x.

$$7x^{2}+20x+8=11$$

Subtract 11 from both sides.

$$7x^{2}+20x+8-11=0$$

Subtract 11 from 8 to get -3.

$$7x^{2}+20x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 20 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 7\left(-3\right)}}{2\times 7}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 7\left(-3\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-20±\sqrt{400-28\left(-3\right)}}{2\times 7}$$

Multiply -28 times -3.

$$x=\frac{-20±\sqrt{400+84}}{2\times 7}$$

Add 400 to 84.

$$x=\frac{-20±\sqrt{484}}{2\times 7}$$

Take the square root of 484.

$$x=\frac{-20±22}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-20±22}{14}$$

Now solve the equation x=\frac{-20±22}{14} when ± is plus. Add -20 to 22.

$$x=\frac{2}{14}$$

Reduce the fraction \frac{2}{14} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{7}$$

Now solve the equation x=\frac{-20±22}{14} when ± is minus. Subtract 22 from -20.

$$x=-\frac{42}{14}$$

Divide -42 by 14.

$$x=-3$$

The equation is now solved.

$$x=\frac{1}{7} x=-3$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$10x^{2}+10x+8-3x^{2}=-10x+11$$

Combine 10x^{2} and -3x^{2} to get 7x^{2}.

$$7x^{2}+10x+8=-10x+11$$

Add 10x to both sides.

$$7x^{2}+10x+8+10x=11$$

Combine 10x and 10x to get 20x.

$$7x^{2}+20x+8=11$$

Subtract 8 from both sides.

$$7x^{2}+20x=11-8$$

Subtract 8 from 11 to get 3.

$$7x^{2}+20x=3$$

Divide both sides by 7.

$$\frac{7x^{2}+20x}{7}=\frac{3}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}+\frac{20}{7}x=\frac{3}{7}$$

Divide \frac{20}{7}, the coefficient of the x term, by 2 to get \frac{10}{7}. Then add the square of \frac{10}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{20}{7}x+\left(\frac{10}{7}\right)^{2}=\frac{3}{7}+\left(\frac{10}{7}\right)^{2}$$

Square \frac{10}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{20}{7}x+\frac{100}{49}=\frac{3}{7}+\frac{100}{49}$$

Add \frac{3}{7} to \frac{100}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{20}{7}x+\frac{100}{49}=\frac{121}{49}$$

Factor x^{2}+\frac{20}{7}x+\frac{100}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{10}{7}\right)^{2}=\frac{121}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{10}{7}\right)^{2}}=\sqrt{\frac{121}{49}}$$

Simplify.

$$x+\frac{10}{7}=\frac{11}{7} x+\frac{10}{7}=-\frac{11}{7}$$

Subtract \frac{10}{7} from both sides of the equation.

$$x=\frac{1}{7} x=-3$$