3q^2-300q+2500=0

Solution for 3q^2-300q+2500=0 equation:

3q^2-300q+2500=0

a = 3; b = -300; c = +2500;
Δ = b2-4ac
Δ = -3002-4·3·2500
Δ = 60000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{60000}=\sqrt{10000*6}=\sqrt{10000}*\sqrt{6}=100\sqrt{6}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-300)-100\sqrt{6}}{2*3}=\frac{300-100\sqrt{6}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-300)+100\sqrt{6}}{2*3}=\frac{300+100\sqrt{6}}{6}
$