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4a^2+4a-120=0
a = 4; b = 4; c = -120;
Δ = b2-4ac
Δ = 42-4·4·(-120)
Δ = 1936
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{1936}=44$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-44}{2*4}=\frac{-48}{8} =-6 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+44}{2*4}=\frac{40}{8} =5 $
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