40x^2+37x+4=0

Solution for 40x^2+37x+4=0 equation:

40x^2+37x+4=0

a = 40; b = 37; c = +4;
Δ = b2-4ac
Δ = 372-4·40·4
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-27}{2*40}=\frac{-64}{80}
=-4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+27}{2*40}=\frac{-10}{80}
=-1/8
$