Solve 2x^2+11x+9=10x+10

$$2 x ^ { 2 } + 11 x + 9 = 10 x + 10$$

Solve for x

$$x=-1$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Subtract 10x from both sides.

$$2x^{2}+11x+9-10x=10$$

Combine 11x and -10x to get x.

$$2x^{2}+x+9=10$$

Subtract 10 from both sides.

$$2x^{2}+x+9-10=0$$

Subtract 10 from 9 to get -1.

$$2x^{2}+x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

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

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. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite 2x^{2}+x-1 as \left(2x^{2}-x\right)+\left(2x-1\right).

$$\left(2x^{2}-x\right)+\left(2x-1\right)$$

Factor out x in 2x^{2}-x.

$$x\left(2x-1\right)+2x-1$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(x+1\right)$$

To find equation solutions, solve 2x-1=0 and x+1=0.

$$x=\frac{1}{2} x=-1$$

Steps Using the Quadratic Formula

Subtract 10x from both sides.

$$2x^{2}+11x+9-10x=10$$

Combine 11x and -10x to get x.

$$2x^{2}+x+9=10$$

Subtract 10 from both sides.

$$2x^{2}+x+9-10=0$$

Subtract 10 from 9 to get -1.

$$2x^{2}+x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-1±\sqrt{1+8}}{2\times 2}$$

Add 1 to 8.

$$x=\frac{-1±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-1±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±3}{4}$$

Now solve the equation x=\frac{-1±3}{4} when ± is plus. Add -1 to 3.

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

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

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-1±3}{4} when ± is minus. Subtract 3 from -1.

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

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

$$x=\frac{1}{2} x=-1$$

Steps for Completing the Square

Subtract 10x from both sides.

$$2x^{2}+11x+9-10x=10$$

Combine 11x and -10x to get x.

$$2x^{2}+x+9=10$$

Subtract 9 from both sides.

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

Subtract 9 from 10 to get 1.

$$2x^{2}+x=1$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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.

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

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

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

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{3}{4} x+\frac{1}{4}=-\frac{3}{4}$$

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

$$x=\frac{1}{2} x=-1$$