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4a^2+16a-20=0
a = 4; b = 16; c = -20;
Δ = b2-4ac
Δ = 162-4·4·(-20)
Δ = 576
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{576}=24$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24}{2*4}=\frac{-40}{8} =-5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24}{2*4}=\frac{8}{8} =1 $
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