20x^2+18x-5=0

Solution for 20x^2+18x-5=0 equation:

20x^2+18x-5=0

a = 20; b = 18; c = -5;
Δ = b2-4ac
Δ = 182-4·20·(-5)
Δ = 724
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:

$\sqrt{\Delta}=\sqrt{724}=\sqrt{4*181}=\sqrt{4}*\sqrt{181}=2\sqrt{181}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{181}}{2*20}=\frac{-18-2\sqrt{181}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{181}}{2*20}=\frac{-18+2\sqrt{181}}{40}
$