10q(q+2)+3(q-3)=5

Solution for 10q(q+2)+3(q-3)=5 equation:

10q(q+2)+3(q-3)=5

We move all terms to the left:

10q(q+2)+3(q-3)-(5)=0

We multiply parentheses

10q^2+20q+3q-9-5=0

We add all the numbers together, and all the variables

10q^2+23q-14=0

a = 10; b = 23; c = -14;
Δ = b2-4ac
Δ = 232-4·10·(-14)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-33}{2*10}=\frac{-56}{20}
=-2+4/5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+33}{2*10}=\frac{10}{20}
=1/2
$