10y^2+19y=2

Solution for 10y^2+19y=2 equation:

10y^2+19y=2

We move all terms to the left:

10y^2+19y-(2)=0

a = 10; b = 19; c = -2;
Δ = b2-4ac
Δ = 192-4·10·(-2)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*10}=\frac{-40}{20}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*10}=\frac{2}{20}
=1/10
$