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

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

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

We multiply parentheses

24u^2+18u=0

a = 24; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·24·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*24}=\frac{-36}{48}
=-3/4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*24}=\frac{0}{48}
=0
$