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a^2-20a-96=0
a = 1; b = -20; c = -96;
Δ = b2-4ac
Δ = -202-4·1·(-96)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{784}=28$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-28}{2*1}=\frac{-8}{2} =-4 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+28}{2*1}=\frac{48}{2} =24 $
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