20x^2+2x-9=0

Solution for 20x^2+2x-9=0 equation:

20x^2+2x-9=0

a = 20; b = 2; c = -9;
Δ = b2-4ac
Δ = 22-4·20·(-9)
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{-(2)-2\sqrt{181}}{2*20}=\frac{-2-2\sqrt{181}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{181}}{2*20}=\frac{-2+2\sqrt{181}}{40}
$