(3u+4)(6+u)=0

Solution for (3u+4)(6+u)=0 equation:

(3u+4)(6+u)=0

We add all the numbers together, and all the variables

(3u+4)(u+6)=0

We multiply parentheses ..

(+3u^2+18u+4u+24)=0

We get rid of parentheses

3u^2+18u+4u+24=0

We add all the numbers together, and all the variables

3u^2+22u+24=0

a = 3; b = 22; c = +24;
Δ = b2-4ac
Δ = 222-4·3·24
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{-(22)-14}{2*3}=\frac{-36}{6}
=-6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*3}=\frac{-8}{6}
=-1+1/3
$