2x-10+3x-20+x^2=108

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Solution for 2x-10+3x-20+x^2=108 equation:



2x-10+3x-20+x^2=108
We move all terms to the left:
2x-10+3x-20+x^2-(108)=0
We add all the numbers together, and all the variables
x^2+5x-138=0
a = 1; b = 5; c = -138;
Δ = b2-4ac
Δ = 52-4·1·(-138)
Δ = 577
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{-(5)-\sqrt{577}}{2*1}=\frac{-5-\sqrt{577}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{577}}{2*1}=\frac{-5+\sqrt{577}}{2} $

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