a^2+20a-65=0

Solution for a^2+20a-65=0 equation:

a^2+20a-65=0

a = 1; b = 20; c = -65;
Δ = b2-4ac
Δ = 202-4·1·(-65)
Δ = 660
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{660}=\sqrt{4*165}=\sqrt{4}*\sqrt{165}=2\sqrt{165}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{165}}{2*1}=\frac{-20-2\sqrt{165}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{165}}{2*1}=\frac{-20+2\sqrt{165}}{2}
$