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7x^2+14x-35=0
a = 7; b = 14; c = -35;
Δ = b2-4ac
Δ = 142-4·7·(-35)
Δ = 1176
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1176}=\sqrt{196*6}=\sqrt{196}*\sqrt{6}=14\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{6}}{2*7}=\frac{-14-14\sqrt{6}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{6}}{2*7}=\frac{-14+14\sqrt{6}}{14} $
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