k2+32k=2880

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Solution for k2+32k=2880 equation:



k2+32k=2880
We move all terms to the left:
k2+32k-(2880)=0
We add all the numbers together, and all the variables
k^2+32k-2880=0
a = 1; b = 32; c = -2880;
Δ = b2-4ac
Δ = 322-4·1·(-2880)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-112}{2*1}=\frac{-144}{2} =-72 $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+112}{2*1}=\frac{80}{2} =40 $

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