(b(b+3))/2=20

Solution for (b(b+3))/2=20 equation:

(b(b+3))/2=20

We move all terms to the left:

(b(b+3))/2-(20)=0

We multiply all the terms by the denominator


(b(b+3))-20*2=0


We calculate terms in parentheses: +(b(b+3)), so:

b(b+3)

We multiply parentheses

b^2+3b

Back to the equation:

+(b^2+3b)


We add all the numbers together, and all the variables

(b^2+3b)-40=0

We get rid of parentheses

b^2+3b-40=0

a = 1; b = 3; c = -40;
Δ = b2-4ac
Δ = 32-4·1·(-40)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*1}=\frac{-16}{2}
=-8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*1}=\frac{10}{2}
=5
$