3f^2+14f+15=0

Solution for 3f^2+14f+15=0 equation:

3f^2+14f+15=0

a = 3; b = 14; c = +15;
Δ = b2-4ac
Δ = 142-4·3·15
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*3}=\frac{-18}{6}
=-3
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*3}=\frac{-10}{6}
=-1+2/3
$