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4a^2+8a-140=0
a = 4; b = 8; c = -140;
Δ = b2-4ac
Δ = 82-4·4·(-140)
Δ = 2304
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{2304}=48$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-48}{2*4}=\frac{-56}{8} =-7 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+48}{2*4}=\frac{40}{8} =5 $
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