Solve 2x^2+2x-5=x^2+61-3x

$$2 x ^ { 2 } + 2 x - 5 = x ^ { 2 } + 61 - 3 x$$

Solve for x

$$x=-11$$

$$x=6$$

Steps Using Factoring

Subtract x^{2} from both sides.

$$2x^{2}+2x-5-x^{2}=61-3x$$

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

$$x^{2}+2x-5=61-3x$$

Subtract 61 from both sides.

$$x^{2}+2x-5-61=-3x$$

Subtract 61 from -5 to get -66.

$$x^{2}+2x-66=-3x$$

Add 3x to both sides.

$$x^{2}+2x-66+3x=0$$

Combine 2x and 3x to get 5x.

$$x^{2}+5x-66=0$$

To solve the equation, factor x^{2}+5x-66 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=5 ab=-66$$

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 -66.

$$-1,66 -2,33 -3,22 -6,11$$

Calculate the sum for each pair.

$$-1+66=65 -2+33=31 -3+22=19 -6+11=5$$

The solution is the pair that gives sum 5.

$$a=-6 b=11$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-6\right)\left(x+11\right)$$

To find equation solutions, solve x-6=0 and x+11=0.

$$x=6 x=-11$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$2x^{2}+2x-5-x^{2}=61-3x$$

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

$$x^{2}+2x-5=61-3x$$

Subtract 61 from both sides.

$$x^{2}+2x-5-61=-3x$$

Subtract 61 from -5 to get -66.

$$x^{2}+2x-66=-3x$$

Add 3x to both sides.

$$x^{2}+2x-66+3x=0$$

Combine 2x and 3x to get 5x.

$$x^{2}+5x-66=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-66. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-66\right)=-66$$

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 -66.

$$-1,66 -2,33 -3,22 -6,11$$

Calculate the sum for each pair.

$$-1+66=65 -2+33=31 -3+22=19 -6+11=5$$

The solution is the pair that gives sum 5.

$$a=-6 b=11$$

Rewrite x^{2}+5x-66 as \left(x^{2}-6x\right)+\left(11x-66\right).

$$\left(x^{2}-6x\right)+\left(11x-66\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(x-6\right)+11\left(x-6\right)$$

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

$$\left(x-6\right)\left(x+11\right)$$

To find equation solutions, solve x-6=0 and x+11=0.

$$x=6 x=-11$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$2x^{2}+2x-5-x^{2}=61-3x$$

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

$$x^{2}+2x-5=61-3x$$

Subtract 61 from both sides.

$$x^{2}+2x-5-61=-3x$$

Subtract 61 from -5 to get -66.

$$x^{2}+2x-66=-3x$$

Add 3x to both sides.

$$x^{2}+2x-66+3x=0$$

Combine 2x and 3x to get 5x.

$$x^{2}+5x-66=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -66 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-66\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-66\right)}}{2}$$

Multiply -4 times -66.

$$x=\frac{-5±\sqrt{25+264}}{2}$$

Add 25 to 264.

$$x=\frac{-5±\sqrt{289}}{2}$$

Take the square root of 289.

$$x=\frac{-5±17}{2}$$

Now solve the equation x=\frac{-5±17}{2} when ± is plus. Add -5 to 17.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{-5±17}{2} when ± is minus. Subtract 17 from -5.

$$x=-\frac{22}{2}$$

Divide -22 by 2.

$$x=-11$$

The equation is now solved.

$$x=6 x=-11$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$2x^{2}+2x-5-x^{2}=61-3x$$

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

$$x^{2}+2x-5=61-3x$$

Add 3x to both sides.

$$x^{2}+2x-5+3x=61$$

Combine 2x and 3x to get 5x.

$$x^{2}+5x-5=61$$

Add 5 to both sides.

$$x^{2}+5x=61+5$$

Add 61 and 5 to get 66.

$$x^{2}+5x=66$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=66+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=66+\frac{25}{4}$$

Add 66 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{289}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{17}{2} x+\frac{5}{2}=-\frac{17}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=6 x=-11$$