2x^2+39x-20=0

Solution for 2x^2+39x-20=0 equation:

2x^2+39x-20=0

a = 2; b = 39; c = -20;
Δ = b2-4ac
Δ = 392-4·2·(-20)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-41}{2*2}=\frac{-80}{4}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+41}{2*2}=\frac{2}{4}
=1/2
$