4f^2+10f+4=0

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Solution for 4f^2+10f+4=0 equation: 



4f^2+10f+4=0

a = 4; b = 10; c = +4;
Δ = b2-4ac
Δ = 102-4·4·4
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*4}=\frac{-16}{8}
=-2
$


$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*4}=\frac{-4}{8}
=-1/2
$

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