11f^2+24f+4=0

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Solution for 11f^2+24f+4=0 equation:



11f^2+24f+4=0
a = 11; b = 24; c = +4;
Δ = b2-4ac
Δ = 242-4·11·4
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-20}{2*11}=\frac{-44}{22} =-2 $
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+20}{2*11}=\frac{-4}{22} =-2/11 $

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