3f^2+26f+16=0

Solution for 3f^2+26f+16=0 equation:

3f^2+26f+16=0

a = 3; b = 26; c = +16;
Δ = b2-4ac
Δ = 262-4·3·16
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-22}{2*3}=\frac{-48}{6}
=-8
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+22}{2*3}=\frac{-4}{6}
=-2/3
$