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3x^2+14x-960=0
a = 3; b = 14; c = -960;
Δ = b2-4ac
Δ = 142-4·3·(-960)
Δ = 11716
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11716}=\sqrt{4*2929}=\sqrt{4}*\sqrt{2929}=2\sqrt{2929}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{2929}}{2*3}=\frac{-14-2\sqrt{2929}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{2929}}{2*3}=\frac{-14+2\sqrt{2929}}{6} $
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