b^2+300b-90000=0

Solution for b^2+300b-90000=0 equation:

b^2+300b-90000=0

a = 1; b = 300; c = -90000;
Δ = b2-4ac
Δ = 3002-4·1·(-90000)
Δ = 450000
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:

$\sqrt{\Delta}=\sqrt{450000}=\sqrt{90000*5}=\sqrt{90000}*\sqrt{5}=300\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(300)-300\sqrt{5}}{2*1}=\frac{-300-300\sqrt{5}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(300)+300\sqrt{5}}{2*1}=\frac{-300+300\sqrt{5}}{2}
$