Q^2-3q^2-25q+75=0

Solution for Q^2-3q^2-25q+75=0 equation:

^2-3Q^2-25Q+75=0

We add all the numbers together, and all the variables

-3Q^2-25Q=0

a = -3; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·(-3)·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*-3}=\frac{0}{-6}
=0
$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*-3}=\frac{50}{-6}
=-8+1/3
$