68=4+b(b+2)

Solution for 68=4+b(b+2) equation:

68=4+b(b+2)

We move all terms to the left:

68-(4+b(b+2))=0


We calculate terms in parentheses: -(4+b(b+2)), so:

4+b(b+2)

determiningTheFunctionDomain
b(b+2)+4

We multiply parentheses

b^2+2b+4

Back to the equation:

-(b^2+2b+4)


We get rid of parentheses

-b^2-2b-4+68=0

We add all the numbers together, and all the variables

-1b^2-2b+64=0

a = -1; b = -2; c = +64;
Δ = b2-4ac
Δ = -22-4·(-1)·64
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{65}}{2*-1}=\frac{2-2\sqrt{65}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{65}}{2*-1}=\frac{2+2\sqrt{65}}{-2}
$