5u^2+3+16u=0

Solution for 5u^2+3+16u=0 equation:

5u^2+3+16u=0

a = 5; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·5·3
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*5}=\frac{-30}{10}
=-3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*5}=\frac{-2}{10}
=-1/5
$