2X^2-42x+44=0

Solution for 2X^2-42x+44=0 equation:

2X^2-42X+44=0

a = 2; b = -42; c = +44;
Δ = b2-4ac
Δ = -422-4·2·44
Δ = 1412
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:

$\sqrt{\Delta}=\sqrt{1412}=\sqrt{4*353}=\sqrt{4}*\sqrt{353}=2\sqrt{353}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{353}}{2*2}=\frac{42-2\sqrt{353}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{353}}{2*2}=\frac{42+2\sqrt{353}}{4}
$