3u^2+17u+20=0

Solution for 3u^2+17u+20=0 equation:

3u^2+17u+20=0

a = 3; b = 17; c = +20;
Δ = b2-4ac
Δ = 172-4·3·20
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*3}=\frac{-24}{6}
=-4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*3}=\frac{-10}{6}
=-1+2/3
$