x^2+20x-35=0

Solution for x^2+20x-35=0 equation:

x^2+20x-35=0

a = 1; b = 20; c = -35;
Δ = b2-4ac
Δ = 202-4·1·(-35)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6\sqrt{15}}{2*1}=\frac{-20-6\sqrt{15}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6\sqrt{15}}{2*1}=\frac{-20+6\sqrt{15}}{2}
$