40x^2+53x+9=0

Solution for 40x^2+53x+9=0 equation:

40x^2+53x+9=0

a = 40; b = 53; c = +9;
Δ = b2-4ac
Δ = 532-4·40·9
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-37}{2*40}=\frac{-90}{80}
=-1+1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+37}{2*40}=\frac{-16}{80}
=-1/5
$