10x^2+2x-20=0

Solution for 10x^2+2x-20=0 equation:

10x^2+2x-20=0

a = 10; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·10·(-20)
Δ = 804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{201}}{2*10}=\frac{-2-2\sqrt{201}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{201}}{2*10}=\frac{-2+2\sqrt{201}}{20}
$