6y^2+10=19y

Solution for 6y^2+10=19y equation:

6y^2+10=19y

We move all terms to the left:

6y^2+10-(19y)=0

a = 6; b = -19; c = +10;
Δ = b2-4ac
Δ = -192-4·6·10
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*6}=\frac{8}{12}
=2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*6}=\frac{30}{12}
=2+1/2
$