(b+4)(2b-)=0

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

(b+4)(2b-)=0

We add all the numbers together, and all the variables

(b+4)(+2b)=0

We multiply parentheses ..

(+2b^2+8b)=0

We get rid of parentheses

2b^2+8b=0

a = 2; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*2}=\frac{-16}{4}
=-4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*2}=\frac{0}{4}
=0
$