3q(4q-7)=0

Solution for 3q(4q-7)=0 equation:

3q(4q-7)=0

We multiply parentheses

12q^2-21q=0

a = 12; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·12·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*12}=\frac{0}{24}
=0
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*12}=\frac{42}{24}
=1+3/4
$