49=7b+b^2

Solution for 49=7b+b^2 equation:

49=7b+b^2

We move all terms to the left:

49-(7b+b^2)=0

We get rid of parentheses

-b^2-7b+49=0

We add all the numbers together, and all the variables

-1b^2-7b+49=0

a = -1; b = -7; c = +49;
Δ = b2-4ac
Δ = -72-4·(-1)·49
Δ = 245
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:

$\sqrt{\Delta}=\sqrt{245}=\sqrt{49*5}=\sqrt{49}*\sqrt{5}=7\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7\sqrt{5}}{2*-1}=\frac{7-7\sqrt{5}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7\sqrt{5}}{2*-1}=\frac{7+7\sqrt{5}}{-2}
$