3s+s^2-40=0

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Solution for 3s+s^2-40=0 equation: 



3s+s^2-40=0

a = 1; b = 3; c = -40;
Δ = b2-4ac
Δ = 32-4·1·(-40)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$


$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*1}=\frac{-16}{2}
=-8
$


$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*1}=\frac{10}{2}
=5
$

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