(u+7)(u+3)=5

Solution for (u+7)(u+3)=5 equation:

(u+7)(u+3)=5

We move all terms to the left:

(u+7)(u+3)-(5)=0

We multiply parentheses ..

(+u^2+3u+7u+21)-5=0

We get rid of parentheses

u^2+3u+7u+21-5=0

We add all the numbers together, and all the variables

u^2+10u+16=0

a = 1; b = 10; c = +16;
Δ = b2-4ac
Δ = 102-4·1·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*1}=\frac{-16}{2}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*1}=\frac{-4}{2}
=-2
$