15y2+14=41y

Solution for 15y2+14=41y equation:

15y^2+14=41y

We move all terms to the left:

15y^2+14-(41y)=0

a = 15; b = -41; c = +14;
Δ = b2-4ac
Δ = -412-4·15·14
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-29}{2*15}=\frac{12}{30}
=2/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+29}{2*15}=\frac{70}{30}
=2+1/3
$