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x^{2}-x-16=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{-\left(-1\right)±\sqrt{1-4\left(-16\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{-\left(-1\right)±\sqrt{1+64}}{2}
Multiply -4 times -16.
x=\frac{-\left(-1\right)±\sqrt{65}}{2}
Add 1 to 64.
x=\frac{1±\sqrt{65}}{2}
The opposite of -1 is 1.
x=\frac{\sqrt{65}+1}{2}
Now solve the equation x=\frac{1±\sqrt{65}}{2} when ± is plus. Add 1 to \sqrt{65}.
x=\frac{1-\sqrt{65}}{2}
Now solve the equation x=\frac{1±\sqrt{65}}{2} when ± is minus. Subtract \sqrt{65} from 1.
x^{2}-x-16=\left(x-\frac{\sqrt{65}+1}{2}\right)\left(x-\frac{1-\sqrt{65}}{2}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1+\sqrt{65}}{2} for x_{1} and \frac{1-\sqrt{65}}{2} for x_{2}.