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5x^2-14x-98=0
a = 5; b = -14; c = -98;
Δ = b2-4ac
Δ = -142-4·5·(-98)
Δ = 2156
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2156}=\sqrt{196*11}=\sqrt{196}*\sqrt{11}=14\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14\sqrt{11}}{2*5}=\frac{14-14\sqrt{11}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14\sqrt{11}}{2*5}=\frac{14+14\sqrt{11}}{10} $
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