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10x^{2}-5x-14=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 10\left(-14\right)}}{2\times 10}
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(-5\right)±\sqrt{25-4\times 10\left(-14\right)}}{2\times 10}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-40\left(-14\right)}}{2\times 10}
Multiply -4 times 10.
x=\frac{-\left(-5\right)±\sqrt{25+560}}{2\times 10}
Multiply -40 times -14.
x=\frac{-\left(-5\right)±\sqrt{585}}{2\times 10}
Add 25 to 560.
x=\frac{-\left(-5\right)±3\sqrt{65}}{2\times 10}
Take the square root of 585.
x=\frac{5±3\sqrt{65}}{2\times 10}
The opposite of -5 is 5.
x=\frac{5±3\sqrt{65}}{20}
Multiply 2 times 10.
x=\frac{3\sqrt{65}+5}{20}
Now solve the equation x=\frac{5±3\sqrt{65}}{20} when ± is plus. Add 5 to 3\sqrt{65}.
x=\frac{3\sqrt{65}}{20}+\frac{1}{4}
Divide 5+3\sqrt{65} by 20.
x=\frac{5-3\sqrt{65}}{20}
Now solve the equation x=\frac{5±3\sqrt{65}}{20} when ± is minus. Subtract 3\sqrt{65} from 5.
x=-\frac{3\sqrt{65}}{20}+\frac{1}{4}
Divide 5-3\sqrt{65} by 20.
10x^{2}-5x-14=10\left(x-\left(\frac{3\sqrt{65}}{20}+\frac{1}{4}\right)\right)\left(x-\left(-\frac{3\sqrt{65}}{20}+\frac{1}{4}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{4}+\frac{3\sqrt{65}}{20} for x_{1} and \frac{1}{4}-\frac{3\sqrt{65}}{20} for x_{2}.