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x^{2}+6x-5=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{-6±\sqrt{6^{2}-4\left(-5\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{-6±\sqrt{36-4\left(-5\right)}}{2}
Square 6.
x=\frac{-6±\sqrt{36+20}}{2}
Multiply -4 times -5.
x=\frac{-6±\sqrt{56}}{2}
Add 36 to 20.
x=\frac{-6±2\sqrt{14}}{2}
Take the square root of 56.
x=\frac{2\sqrt{14}-6}{2}
Now solve the equation x=\frac{-6±2\sqrt{14}}{2} when ± is plus. Add -6 to 2\sqrt{14}.
x=\sqrt{14}-3
Divide -6+2\sqrt{14} by 2.
x=\frac{-2\sqrt{14}-6}{2}
Now solve the equation x=\frac{-6±2\sqrt{14}}{2} when ± is minus. Subtract 2\sqrt{14} from -6.
x=-\sqrt{14}-3
Divide -6-2\sqrt{14} by 2.
x^{2}+6x-5=\left(x-\left(\sqrt{14}-3\right)\right)\left(x-\left(-\sqrt{14}-3\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\sqrt{14} for x_{1} and -3-\sqrt{14} for x_{2}.