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

Solution for (2+u)(4u-1)=0 equation:

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

We add all the numbers together, and all the variables

(u+2)(4u-1)=0

We multiply parentheses ..

(+4u^2-1u+8u-2)=0

We get rid of parentheses

4u^2-1u+8u-2=0

We add all the numbers together, and all the variables

4u^2+7u-2=0

a = 4; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·4·(-2)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*4}=\frac{-16}{8}
=-2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*4}=\frac{2}{8}
=1/4
$