20x^2+4x-81=0

Solution for 20x^2+4x-81=0 equation:

20x^2+4x-81=0

a = 20; b = 4; c = -81;
Δ = b2-4ac
Δ = 42-4·20·(-81)
Δ = 6496
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:

$\sqrt{\Delta}=\sqrt{6496}=\sqrt{16*406}=\sqrt{16}*\sqrt{406}=4\sqrt{406}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{406}}{2*20}=\frac{-4-4\sqrt{406}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{406}}{2*20}=\frac{-4+4\sqrt{406}}{40}
$