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2x^2+56x-93=0
a = 2; b = 56; c = -93;
Δ = b2-4ac
Δ = 562-4·2·(-93)
Δ = 3880
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3880}=\sqrt{4*970}=\sqrt{4}*\sqrt{970}=2\sqrt{970}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-2\sqrt{970}}{2*2}=\frac{-56-2\sqrt{970}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+2\sqrt{970}}{2*2}=\frac{-56+2\sqrt{970}}{4} $
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