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9a^2+36a+20=0
a = 9; b = 36; c = +20;
Δ = b2-4ac
Δ = 362-4·9·20
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$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-24}{2*9}=\frac{-60}{18} =-3+1/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+24}{2*9}=\frac{-12}{18} =-2/3 $
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