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a+b=2 ab=1\left(-3\right)=-3
Factor the expression by grouping. First, the expression 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^{2}+2x-3=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-2±\sqrt{2^{2}-4\left(-3\right)}}{2}
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=\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^{2}+2x-3=\left(x-1\right)\left(x-\left(-3\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -3 for x_{2}.
x^{2}+2x-3=\left(x-1\right)\left(x+3\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.