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

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

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

We multiply parentheses ..

(+6n^2+8n-15n-20)=0

We get rid of parentheses

6n^2+8n-15n-20=0

We add all the numbers together, and all the variables

6n^2-7n-20=0

a = 6; b = -7; c = -20;
Δ = b2-4ac
Δ = -72-4·6·(-20)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-23}{2*6}=\frac{-16}{12}
=-1+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+23}{2*6}=\frac{30}{12}
=2+1/2
$