X^2-42x+196=0

Solution for X^2-42x+196=0 equation:

X^2-42X+196=0

a = 1; b = -42; c = +196;
Δ = b2-4ac
Δ = -422-4·1·196
Δ = 980
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:

$\sqrt{\Delta}=\sqrt{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14\sqrt{5}}{2*1}=\frac{42-14\sqrt{5}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{5}}{2*1}=\frac{42+14\sqrt{5}}{2}
$