b^2-5b-27=-3

Solution for b^2-5b-27=-3 equation:

b^2-5b-27=-3

We move all terms to the left:

b^2-5b-27-(-3)=0

We add all the numbers together, and all the variables

b^2-5b-24=0

a = 1; b = -5; c = -24;
Δ = b2-4ac
Δ = -52-4·1·(-24)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*1}=\frac{-6}{2}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*1}=\frac{16}{2}
=8
$