q^2+60q-7200=0

Solution for q^2+60q-7200=0 equation:

q^2+60q-7200=0

a = 1; b = 60; c = -7200;
Δ = b2-4ac
Δ = 602-4·1·(-7200)
Δ = 32400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{32400}=180$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-180}{2*1}=\frac{-240}{2}
=-120
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+180}{2*1}=\frac{120}{2}
=60
$