180=(5z-3)(4z+3)

Solution for 180=(5z-3)(4z+3) equation:

180=(5z-3)(4z+3)

We move all terms to the left:

180-((5z-3)(4z+3))=0

We multiply parentheses ..

-((+20z^2+15z-12z-9))+180=0


We calculate terms in parentheses: -((+20z^2+15z-12z-9)), so:

(+20z^2+15z-12z-9)

We get rid of parentheses

20z^2+15z-12z-9

We add all the numbers together, and all the variables

20z^2+3z-9

Back to the equation:

-(20z^2+3z-9)


We get rid of parentheses

-20z^2-3z+9+180=0

We add all the numbers together, and all the variables

-20z^2-3z+189=0

a = -20; b = -3; c = +189;
Δ = b2-4ac
Δ = -32-4·(-20)·189
Δ = 15129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15129}=123$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-123}{2*-20}=\frac{-120}{-40}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+123}{2*-20}=\frac{126}{-40}
=-3+3/20
$