b^2-4b-67=10

Solution for b^2-4b-67=10 equation:

b^2-4b-67=10

We move all terms to the left:

b^2-4b-67-(10)=0

We add all the numbers together, and all the variables

b^2-4b-77=0

a = 1; b = -4; c = -77;
Δ = b2-4ac
Δ = -42-4·1·(-77)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-18}{2*1}=\frac{-14}{2}
=-7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+18}{2*1}=\frac{22}{2}
=11
$