49x^2-42x+4=0

Solution for 49x^2-42x+4=0 equation:

49x^2-42x+4=0

a = 49; b = -42; c = +4;
Δ = b2-4ac
Δ = -422-4·49·4
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*49}=\frac{42-14\sqrt{5}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{5}}{2*49}=\frac{42+14\sqrt{5}}{98}
$