4y^2-41y+10=0

Solution for 4y^2-41y+10=0 equation:

4y^2-41y+10=0

a = 4; b = -41; c = +10;
Δ = b2-4ac
Δ = -412-4·4·10
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-39}{2*4}=\frac{2}{8}
=1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+39}{2*4}=\frac{80}{8}
=10
$