18y^2=41y+10

Solution for 18y^2=41y+10 equation:

18y^2=41y+10

We move all terms to the left:

18y^2-(41y+10)=0

We get rid of parentheses

18y^2-41y-10=0

a = 18; b = -41; c = -10;
Δ = b2-4ac
Δ = -412-4·18·(-10)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-49}{2*18}=\frac{-8}{36}
=-2/9
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+49}{2*18}=\frac{90}{36}
=2+1/2
$