42y^2+397y+925=0

Solution for 42y^2+397y+925=0 equation:

42y^2+397y+925=0

a = 42; b = 397; c = +925;
Δ = b2-4ac
Δ = 3972-4·42·925
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(397)-47}{2*42}=\frac{-444}{84}
=-5+2/7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(397)+47}{2*42}=\frac{-350}{84}
=-4+1/6
$