6x^2+42x-98=0

Solution for 6x^2+42x-98=0 equation:

6x^2+42x-98=0

a = 6; b = 42; c = -98;
Δ = b2-4ac
Δ = 422-4·6·(-98)
Δ = 4116
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:

$\sqrt{\Delta}=\sqrt{4116}=\sqrt{196*21}=\sqrt{196}*\sqrt{21}=14\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-14\sqrt{21}}{2*6}=\frac{-42-14\sqrt{21}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+14\sqrt{21}}{2*6}=\frac{-42+14\sqrt{21}}{12}
$