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

Solution for (2+u)(2u-3)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+2u^2-3u+4u-6)=0

We get rid of parentheses

2u^2-3u+4u-6=0

We add all the numbers together, and all the variables

2u^2+u-6=0

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