280q-3q^2-200+10q=0

Solution for 280q-3q^2-200+10q=0 equation:

280q-3q^2-200+10q=0

We add all the numbers together, and all the variables

-3q^2+290q-200=0

a = -3; b = 290; c = -200;
Δ = b2-4ac
Δ = 2902-4·(-3)·(-200)
Δ = 81700
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:

$\sqrt{\Delta}=\sqrt{81700}=\sqrt{100*817}=\sqrt{100}*\sqrt{817}=10\sqrt{817}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(290)-10\sqrt{817}}{2*-3}=\frac{-290-10\sqrt{817}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(290)+10\sqrt{817}}{2*-3}=\frac{-290+10\sqrt{817}}{-6}
$