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20a^2+41a-9=0
a = 20; b = 41; c = -9;
Δ = b2-4ac
Δ = 412-4·20·(-9)
Δ = 2401
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{2401}=49$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-49}{2*20}=\frac{-90}{40} =-2+1/4 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+49}{2*20}=\frac{8}{40} =1/5 $
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