X^2+41x+20=0

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Solution for X^2+41x+20=0 equation: 



X^2+41X+20=0

a = 1; b = 41; c = +20;
Δ = b2-4ac
Δ = 412-4·1·20
Δ = 1601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-\sqrt{1601}}{2*1}=\frac{-41-\sqrt{1601}}{2}
$


$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{1601}}{2*1}=\frac{-41+\sqrt{1601}}{2}
$

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