b^2+b=7b+14

Solution for b^2+b=7b+14 equation:

b^2+b=7b+14

We move all terms to the left:

b^2+b-(7b+14)=0

We get rid of parentheses

b^2+b-7b-14=0

We add all the numbers together, and all the variables

b^2-6b-14=0

a = 1; b = -6; c = -14;
Δ = b2-4ac
Δ = -62-4·1·(-14)
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{23}}{2*1}=\frac{6-2\sqrt{23}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{23}}{2*1}=\frac{6+2\sqrt{23}}{2}
$