3x+(2x+40)(4x-40)=180

Solution for 3x+(2x+40)(4x-40)=180 equation:

3x+(2x+40)(4x-40)=180

We move all terms to the left:

3x+(2x+40)(4x-40)-(180)=0

We multiply parentheses ..

(+8x^2-80x+160x-1600)+3x-180=0

We get rid of parentheses

8x^2-80x+160x+3x-1600-180=0

We add all the numbers together, and all the variables

8x^2+83x-1780=0

a = 8; b = 83; c = -1780;
Δ = b2-4ac
Δ = 832-4·8·(-1780)
Δ = 63849
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{-(83)-\sqrt{63849}}{2*8}=\frac{-83-\sqrt{63849}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(83)+\sqrt{63849}}{2*8}=\frac{-83+\sqrt{63849}}{16}
$