180=65+(15x-4)(4x+5)

Solution for 180=65+(15x-4)(4x+5) equation:

180=65+(15x-4)(4x+5)

We move all terms to the left:

180-(65+(15x-4)(4x+5))=0

We multiply parentheses ..

-(65+(+60x^2+75x-16x-20))+180=0


We calculate terms in parentheses: -(65+(+60x^2+75x-16x-20)), so:

65+(+60x^2+75x-16x-20)

determiningTheFunctionDomain
(+60x^2+75x-16x-20)+65

We get rid of parentheses

60x^2+75x-16x-20+65

We add all the numbers together, and all the variables

60x^2+59x+45

Back to the equation:

-(60x^2+59x+45)


We get rid of parentheses

-60x^2-59x-45+180=0

We add all the numbers together, and all the variables

-60x^2-59x+135=0

a = -60; b = -59; c = +135;
Δ = b2-4ac
Δ = -592-4·(-60)·135
Δ = 35881
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-59)-\sqrt{35881}}{2*-60}=\frac{59-\sqrt{35881}}{-120}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-59)+\sqrt{35881}}{2*-60}=\frac{59+\sqrt{35881}}{-120}
$