4n2+20n-11=0

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Solution for 4n2+20n-11=0 equation:



4n^2+20n-11=0
a = 4; b = 20; c = -11;
Δ = b2-4ac
Δ = 202-4·4·(-11)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-24}{2*4}=\frac{-44}{8} =-5+1/2 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+24}{2*4}=\frac{4}{8} =1/2 $

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