(4q+5)(3q-2)=0

Solution for (4q+5)(3q-2)=0 equation:

(4q+5)(3q-2)=0

We multiply parentheses ..

(+12q^2-8q+15q-10)=0

We get rid of parentheses

12q^2-8q+15q-10=0

We add all the numbers together, and all the variables

12q^2+7q-10=0

a = 12; b = 7; c = -10;
Δ = b2-4ac
Δ = 72-4·12·(-10)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*12}=\frac{-30}{24}
=-1+1/4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*12}=\frac{16}{24}
=2/3
$