b^2+28b-27=23

Solution for b^2+28b-27=23 equation:

b^2+28b-27=23

We move all terms to the left:

b^2+28b-27-(23)=0

We add all the numbers together, and all the variables

b^2+28b-50=0

a = 1; b = 28; c = -50;
Δ = b2-4ac
Δ = 282-4·1·(-50)
Δ = 984
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:

$\sqrt{\Delta}=\sqrt{984}=\sqrt{4*246}=\sqrt{4}*\sqrt{246}=2\sqrt{246}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{246}}{2*1}=\frac{-28-2\sqrt{246}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{246}}{2*1}=\frac{-28+2\sqrt{246}}{2}
$