(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)(-1u+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
$