(3x-40)(2x-5)=180

Solution for (3x-40)(2x-5)=180 equation:

(3x-40)(2x-5)=180

We move all terms to the left:

(3x-40)(2x-5)-(180)=0

We multiply parentheses ..

(+6x^2-15x-80x+200)-180=0

We get rid of parentheses

6x^2-15x-80x+200-180=0

We add all the numbers together, and all the variables

6x^2-95x+20=0

a = 6; b = -95; c = +20;
Δ = b2-4ac
Δ = -952-4·6·20
Δ = 8545
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{-(-95)-\sqrt{8545}}{2*6}=\frac{95-\sqrt{8545}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{8545}}{2*6}=\frac{95+\sqrt{8545}}{12}
$