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9a^2+41a+20=0
a = 9; b = 41; c = +20;
Δ = b2-4ac
Δ = 412-4·9·20
Δ = 961
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{961}=31$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-31}{2*9}=\frac{-72}{18} =-4 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+31}{2*9}=\frac{-10}{18} =-5/9 $
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