3u^2+16u=-5

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

3u^2+16u=-5

We move all terms to the left:

3u^2+16u-(-5)=0

We add all the numbers together, and all the variables

3u^2+16u+5=0

a = 3; b = 16; c = +5;
Δ = b2-4ac
Δ = 162-4·3·5
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*3}=\frac{-30}{6}
=-5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*3}=\frac{-2}{6}
=-1/3
$