.5*b(3b+2)=28^2

Solution for .5*b(3b+2)=28^2 equation:

.5b(3b+2)=28^2

We move all terms to the left:

.5b(3b+2)-(28^2)=0

We add all the numbers together, and all the variables

.5b(3b+2)-784=0

We multiply parentheses

3b^2+2b-784=0

a = 3; b = 2; c = -784;
Δ = b2-4ac
Δ = 22-4·3·(-784)
Δ = 9412
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9412}=\sqrt{4*2353}=\sqrt{4}*\sqrt{2353}=2\sqrt{2353}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2353}}{2*3}=\frac{-2-2\sqrt{2353}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2353}}{2*3}=\frac{-2+2\sqrt{2353}}{6}
$