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

Solution for (4u-2)(6+u)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4u^2+24u-2u-12)=0

We get rid of parentheses

4u^2+24u-2u-12=0

We add all the numbers together, and all the variables

4u^2+22u-12=0

a = 4; b = 22; c = -12;
Δ = b2-4ac
Δ = 222-4·4·(-12)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*4}=\frac{-48}{8}
=-6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*4}=\frac{4}{8}
=1/2
$