4y^2-42y+20=0

Solution for 4y^2-42y+20=0 equation:

4y^2-42y+20=0

a = 4; b = -42; c = +20;
Δ = b2-4ac
Δ = -422-4·4·20
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-38}{2*4}=\frac{4}{8}
=1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+38}{2*4}=\frac{80}{8}
=10
$