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

Solution for (1+u)(3u+4)=0 equation:

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

We add all the numbers together, and all the variables

(u+1)(3u+4)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3u^2+7u+4=0

a = 3; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·3·4
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*3}=\frac{-8}{6}
=-1+1/3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*3}=\frac{-6}{6}
=-1
$