5x2+20x-10=0

Solution for 5x2+20x-10=0 equation:

5x^2+20x-10=0

a = 5; b = 20; c = -10;
Δ = b2-4ac
Δ = 202-4·5·(-10)
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{6}}{2*5}=\frac{-20-10\sqrt{6}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{6}}{2*5}=\frac{-20+10\sqrt{6}}{10}
$