20x^2+40x+9=0

Solution for 20x^2+40x+9=0 equation:

20x^2+40x+9=0

a = 20; b = 40; c = +9;
Δ = b2-4ac
Δ = 402-4·20·9
Δ = 880
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:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{55}}{2*20}=\frac{-40-4\sqrt{55}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{55}}{2*20}=\frac{-40+4\sqrt{55}}{40}
$