3u^2+34u+11=0

Solution for 3u^2+34u+11=0 equation:

3u^2+34u+11=0

a = 3; b = 34; c = +11;
Δ = b2-4ac
Δ = 342-4·3·11
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-32}{2*3}=\frac{-66}{6}
=-11
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+32}{2*3}=\frac{-2}{6}
=-1/3
$