2b^2+7b=b^2+6b+72

Solution for 2b^2+7b=b^2+6b+72 equation:

2b^2+7b=b^2+6b+72

We move all terms to the left:

2b^2+7b-(b^2+6b+72)=0

We get rid of parentheses

2b^2-b^2+7b-6b-72=0

We add all the numbers together, and all the variables

b^2+b-72=0

a = 1; b = 1; c = -72;
Δ = b2-4ac
Δ = 12-4·1·(-72)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*1}=\frac{-18}{2}
=-9
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*1}=\frac{16}{2}
=8
$