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x^{2}+2x+5-8=0
Subtract 8 from both sides.
x^{2}+2x-3=0
Subtract 8 from 5 to get -3.
a+b=2 ab=-3
To solve the equation, factor x^{2}+2x-3 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=-1 b=3
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.
\left(x-1\right)\left(x+3\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=1 x=-3
To find equation solutions, solve x-1=0 and x+3=0.
x^{2}+2x+5-8=0
Subtract 8 from both sides.
x^{2}+2x-3=0
Subtract 8 from 5 to get -3.
a+b=2 ab=1\left(-3\right)=-3
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
a=-1 b=3
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.
\left(x^{2}-x\right)+\left(3x-3\right)
Rewrite x^{2}+2x-3 as \left(x^{2}-x\right)+\left(3x-3\right).
x\left(x-1\right)+3\left(x-1\right)
Factor out x in the first and 3 in the second group.
\left(x-1\right)\left(x+3\right)
Factor out common term x-1 by using distributive property.
x=1 x=-3
To find equation solutions, solve x-1=0 and x+3=0.
x^{2}+2x+5=8
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x^{2}+2x+5-8=8-8
Subtract 8 from both sides of the equation.
x^{2}+2x+5-8=0
Subtracting 8 from itself leaves 0.
x^{2}+2x-3=0
Subtract 8 from 5.
x=\frac{-2±\sqrt{2^{2}-4\left(-3\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\left(-3\right)}}{2}
Square 2.
x=\frac{-2±\sqrt{4+12}}{2}
Multiply -4 times -3.
x=\frac{-2±\sqrt{16}}{2}
Add 4 to 12.
x=\frac{-2±4}{2}
Take the square root of 16.
x=\frac{2}{2}
Now solve the equation x=\frac{-2±4}{2} when ± is plus. Add -2 to 4.
x=1
Divide 2 by 2.
x=-\frac{6}{2}
Now solve the equation x=\frac{-2±4}{2} when ± is minus. Subtract 4 from -2.
x=-3
Divide -6 by 2.
x=1 x=-3
The equation is now solved.
x^{2}+2x+5=8
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
x^{2}+2x+5-5=8-5
Subtract 5 from both sides of the equation.
x^{2}+2x=8-5
Subtracting 5 from itself leaves 0.
x^{2}+2x=3
Subtract 5 from 8.
x^{2}+2x+1^{2}=3+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=3+1
Square 1.
x^{2}+2x+1=4
Add 3 to 1.
\left(x+1\right)^{2}=4
Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+1\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
x+1=2 x+1=-2
Simplify.
x=1 x=-3
Subtract 1 from both sides of the equation.