(u+2)(u-4)=5

Solution for (u+2)(u-4)=5 equation:

(u+2)(u-4)=5

We move all terms to the left:

(u+2)(u-4)-(5)=0

We multiply parentheses ..

(+u^2-4u+2u-8)-5=0

We get rid of parentheses

u^2-4u+2u-8-5=0

We add all the numbers together, and all the variables

u^2-2u-13=0

a = 1; b = -2; c = -13;
Δ = b2-4ac
Δ = -22-4·1·(-13)
Δ = 56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{14}}{2*1}=\frac{2-2\sqrt{14}}{2}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{14}}{2*1}=\frac{2+2\sqrt{14}}{2}
$