B)(x+4)(x+2)-10=0

Solution for B)(x+4)(x+2)-10=0 equation:

)(B+4)(B+2)-10=0

We multiply parentheses ..

)(+B^2+2B+4B+8)-10=0

We add all the numbers together, and all the variables

B^2+2B+4B+)8-10=0

We add all the numbers together, and all the variables

B^2+6B=0

a = 1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*1}=\frac{-12}{2}
=-6
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*1}=\frac{0}{2}
=0
$