(s)2+s=48

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Solution for (s)2+s=48 equation:



(s)2+s=48
We move all terms to the left:
(s)2+s-(48)=0
We add all the numbers together, and all the variables
s^2+s-48=0
a = 1; b = 1; c = -48;
Δ = b2-4ac
Δ = 12-4·1·(-48)
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{193}}{2*1}=\frac{-1-\sqrt{193}}{2} $
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*1}=\frac{-1+\sqrt{193}}{2} $

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