20=0.5b(b+4)

Solution for 20=0.5b(b+4) equation:

20=0.5b(b+4)

We move all terms to the left:

20-(0.5b(b+4))=0


We calculate terms in parentheses: -(0.5b(b+4)), so:

0.5b(b+4)

We multiply parentheses

0b^2+0b

We add all the numbers together, and all the variables

b^2+b

Back to the equation:

-(b^2+b)


We get rid of parentheses

-b^2-b+20=0

We add all the numbers together, and all the variables

-1b^2-1b+20=0

a = -1; b = -1; c = +20;
Δ = b2-4ac
Δ = -12-4·(-1)·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*-1}=\frac{-8}{-2}
=+4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*-1}=\frac{10}{-2}
=-5
$