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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3u^2-1u-4=0

a = 3; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·3·(-4)
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{-(-1)-7}{2*3}=\frac{-6}{6}
=-1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*3}=\frac{8}{6}
=1+1/3
$