2x^2-42x+2=0

Solution for 2x^2-42x+2=0 equation:

2x^2-42x+2=0

a = 2; b = -42; c = +2;
Δ = b2-4ac
Δ = -422-4·2·2
Δ = 1748
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1748}=\sqrt{4*437}=\sqrt{4}*\sqrt{437}=2\sqrt{437}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{437}}{2*2}=\frac{42-2\sqrt{437}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{437}}{2*2}=\frac{42+2\sqrt{437}}{4}
$