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5x^{2}-10x-35=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(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5\left(-35\right)}}{2\times 5}
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(-10\right)±\sqrt{100-4\times 5\left(-35\right)}}{2\times 5}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-20\left(-35\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-10\right)±\sqrt{100+700}}{2\times 5}
Multiply -20 times -35.
x=\frac{-\left(-10\right)±\sqrt{800}}{2\times 5}
Add 100 to 700.
x=\frac{-\left(-10\right)±20\sqrt{2}}{2\times 5}
Take the square root of 800.
x=\frac{10±20\sqrt{2}}{2\times 5}
The opposite of -10 is 10.
x=\frac{10±20\sqrt{2}}{10}
Multiply 2 times 5.
x=\frac{20\sqrt{2}+10}{10}
Now solve the equation x=\frac{10±20\sqrt{2}}{10} when ± is plus. Add 10 to 20\sqrt{2}.
x=2\sqrt{2}+1
Divide 20\sqrt{2}+10 by 10.
x=\frac{10-20\sqrt{2}}{10}
Now solve the equation x=\frac{10±20\sqrt{2}}{10} when ± is minus. Subtract 20\sqrt{2} from 10.
x=1-2\sqrt{2}
Divide 10-20\sqrt{2} by 10.
5x^{2}-10x-35=5\left(x-\left(2\sqrt{2}+1\right)\right)\left(x-\left(1-2\sqrt{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2\sqrt{2}+1 for x_{1} and 1-2\sqrt{2} for x_{2}.