4200=3q^2+15q

Solution for 4200=3q^2+15q equation:

4200=3q^2+15q

We move all terms to the left:

4200-(3q^2+15q)=0

We get rid of parentheses

-3q^2-15q+4200=0

a = -3; b = -15; c = +4200;
Δ = b2-4ac
Δ = -152-4·(-3)·4200
Δ = 50625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{50625}=225$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-225}{2*-3}=\frac{-210}{-6}
=+35
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+225}{2*-3}=\frac{240}{-6}
=-40
$