3u^2+49u+60=0

Solution for 3u^2+49u+60=0 equation:

3u^2+49u+60=0

a = 3; b = 49; c = +60;
Δ = b2-4ac
Δ = 492-4·3·60
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-41}{2*3}=\frac{-90}{6}
=-15
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+41}{2*3}=\frac{-8}{6}
=-1+1/3
$