3x^2-42x+98=0

Solution for 3x^2-42x+98=0 equation:

3x^2-42x+98=0

a = 3; b = -42; c = +98;
Δ = b2-4ac
Δ = -422-4·3·98
Δ = 588
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:

$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14\sqrt{3}}{2*3}=\frac{42-14\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{3}}{2*3}=\frac{42+14\sqrt{3}}{6}
$