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5x^{2}+4x-7=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{-4±\sqrt{4^{2}-4\times 5\left(-7\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{-4±\sqrt{16-4\times 5\left(-7\right)}}{2\times 5}
Square 4.
x=\frac{-4±\sqrt{16-20\left(-7\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-4±\sqrt{16+140}}{2\times 5}
Multiply -20 times -7.
x=\frac{-4±\sqrt{156}}{2\times 5}
Add 16 to 140.
x=\frac{-4±2\sqrt{39}}{2\times 5}
Take the square root of 156.
x=\frac{-4±2\sqrt{39}}{10}
Multiply 2 times 5.
x=\frac{2\sqrt{39}-4}{10}
Now solve the equation x=\frac{-4±2\sqrt{39}}{10} when ± is plus. Add -4 to 2\sqrt{39}.
x=\frac{\sqrt{39}-2}{5}
Divide -4+2\sqrt{39} by 10.
x=\frac{-2\sqrt{39}-4}{10}
Now solve the equation x=\frac{-4±2\sqrt{39}}{10} when ± is minus. Subtract 2\sqrt{39} from -4.
x=\frac{-\sqrt{39}-2}{5}
Divide -4-2\sqrt{39} by 10.
5x^{2}+4x-7=5\left(x-\frac{\sqrt{39}-2}{5}\right)\left(x-\frac{-\sqrt{39}-2}{5}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-2+\sqrt{39}}{5} for x_{1} and \frac{-2-\sqrt{39}}{5} for x_{2}.