20x^2-92x-39=0

Solution for 20x^2-92x-39=0 equation:

20x^2-92x-39=0

a = 20; b = -92; c = -39;
Δ = b2-4ac
Δ = -922-4·20·(-39)
Δ = 11584
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:

$\sqrt{\Delta}=\sqrt{11584}=\sqrt{64*181}=\sqrt{64}*\sqrt{181}=8\sqrt{181}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-92)-8\sqrt{181}}{2*20}=\frac{92-8\sqrt{181}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-92)+8\sqrt{181}}{2*20}=\frac{92+8\sqrt{181}}{40}
$