b^2+18b-819=0

Solution for b^2+18b-819=0 equation:

b^2+18b-819=0

a = 1; b = 18; c = -819;
Δ = b2-4ac
Δ = 182-4·1·(-819)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-60}{2*1}=\frac{-78}{2}
=-39
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+60}{2*1}=\frac{42}{2}
=21
$