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14x^2+8x-35=0
a = 14; b = 8; c = -35;
Δ = b2-4ac
Δ = 82-4·14·(-35)
Δ = 2024
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2024}=\sqrt{4*506}=\sqrt{4}*\sqrt{506}=2\sqrt{506}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{506}}{2*14}=\frac{-8-2\sqrt{506}}{28} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{506}}{2*14}=\frac{-8+2\sqrt{506}}{28} $
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