7(b+3)3b=20

Solution for 7(b+3)3b=20 equation:

7(b+3)3b=20

We move all terms to the left:

7(b+3)3b-(20)=0

We multiply parentheses

21b^2+63b-20=0

a = 21; b = 63; c = -20;
Δ = b2-4ac
Δ = 632-4·21·(-20)
Δ = 5649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-\sqrt{5649}}{2*21}=\frac{-63-\sqrt{5649}}{42}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+\sqrt{5649}}{2*21}=\frac{-63+\sqrt{5649}}{42}
$