2a^2+20a-52=0

Solution for 2a^2+20a-52=0 equation:

2a^2+20a-52=0

a = 2; b = 20; c = -52;
Δ = b2-4ac
Δ = 202-4·2·(-52)
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{51}}{2*2}=\frac{-20-4\sqrt{51}}{4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{51}}{2*2}=\frac{-20+4\sqrt{51}}{4}
$