w2+7w-408=0

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Solution for w2+7w-408=0 equation:



w2+7w-408=0
We add all the numbers together, and all the variables
w^2+7w-408=0
a = 1; b = 7; c = -408;
Δ = b2-4ac
Δ = 72-4·1·(-408)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-41}{2*1}=\frac{-48}{2} =-24 $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+41}{2*1}=\frac{34}{2} =17 $

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