0=u^2-4u-12

Solution for 0=u^2-4u-12 equation:

0=u^2-4u-12

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-u^2+4u+12=0

We add all the numbers together, and all the variables

-1u^2+4u+12=0

a = -1; b = 4; c = +12;
Δ = b2-4ac
Δ = 42-4·(-1)·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-1}=\frac{-12}{-2}
=+6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-1}=\frac{4}{-2}
=-2
$