10w^2-42w+4=0

Solution for 10w^2-42w+4=0 equation:

10w^2-42w+4=0

a = 10; b = -42; c = +4;
Δ = b2-4ac
Δ = -422-4·10·4
Δ = 1604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1604}=\sqrt{4*401}=\sqrt{4}*\sqrt{401}=2\sqrt{401}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{401}}{2*10}=\frac{42-2\sqrt{401}}{20}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{401}}{2*10}=\frac{42+2\sqrt{401}}{20}
$