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20x^2+41x+20=0
a = 20; b = 41; c = +20;
Δ = b2-4ac
Δ = 412-4·20·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-9}{2*20}=\frac{-50}{40} =-1+1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+9}{2*20}=\frac{-32}{40} =-4/5 $
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