3f^2+37f+12=0

Solution for 3f^2+37f+12=0 equation:

3f^2+37f+12=0

a = 3; b = 37; c = +12;
Δ = b2-4ac
Δ = 372-4·3·12
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-35}{2*3}=\frac{-72}{6}
=-12
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+35}{2*3}=\frac{-2}{6}
=-1/3
$