Solve x^2+1x+x^2+2x=5

$${ x }^{ 2 } +1x+ { x }^{ 2 } +2x=5$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=1$$

Steps Using Factoring By Grouping

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

$$2x^{2}+1x+2x=5$$

Combine 1x and 2x to get 3x.

$$2x^{2}+3x=5$$

Subtract 5 from both sides.

$$2x^{2}+3x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\left(-5\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

Rewrite 2x^{2}+3x-5 as \left(2x^{2}-2x\right)+\left(5x-5\right).

$$\left(2x^{2}-2x\right)+\left(5x-5\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(x-1\right)+5\left(x-1\right)$$

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

$$\left(x-1\right)\left(2x+5\right)$$

To find equation solutions, solve x-1=0 and 2x+5=0.

$$x=1 x=-\frac{5}{2}$$

Steps Using the Quadratic Formula

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

$$2x^{2}+1x+2x=5$$

Combine 1x and 2x to get 3x.

$$2x^{2}+3x=5$$

Subtract 5 from both sides.

$$2x^{2}+3x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-3±\sqrt{9+40}}{2\times 2}$$

Add 9 to 40.

$$x=\frac{-3±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$x=\frac{-3±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±7}{4}$$

Now solve the equation x=\frac{-3±7}{4} when ± is plus. Add -3 to 7.

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

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-3±7}{4} when ± is minus. Subtract 7 from -3.

$$x=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{2}$$

The equation is now solved.

$$x=1 x=-\frac{5}{2}$$

Steps for Completing the Square

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

$$2x^{2}+1x+2x=5$$

Combine 1x and 2x to get 3x.

$$2x^{2}+3x=5$$

Divide both sides by 2.

$$\frac{2x^{2}+3x}{2}=\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{3}{2}x=\frac{5}{2}$$

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

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=\frac{5}{2}+\left(\frac{3}{4}\right)^{2}$$

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

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{5}{2}+\frac{9}{16}$$

Add \frac{5}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{49}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{7}{4} x+\frac{3}{4}=-\frac{7}{4}$$

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

$$x=1 x=-\frac{5}{2}$$