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6x^2+14x-98=0
a = 6; b = 14; c = -98;
Δ = b2-4ac
Δ = 142-4·6·(-98)
Δ = 2548
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2548}=\sqrt{196*13}=\sqrt{196}*\sqrt{13}=14\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{13}}{2*6}=\frac{-14-14\sqrt{13}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{13}}{2*6}=\frac{-14+14\sqrt{13}}{12} $
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