20x^2+49x+9=0

Solution for 20x^2+49x+9=0 equation:

20x^2+49x+9=0

a = 20; b = 49; c = +9;
Δ = b2-4ac
Δ = 492-4·20·9
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-41}{2*20}=\frac{-90}{40}
=-2+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+41}{2*20}=\frac{-8}{40}
=-1/5
$