(3x+1)(12x-4)=75

Solution for (3x+1)(12x-4)=75 equation:

(3x+1)(12x-4)=75

We move all terms to the left:

(3x+1)(12x-4)-(75)=0

We multiply parentheses ..

(+36x^2-12x+12x-4)-75=0

We get rid of parentheses

36x^2-12x+12x-4-75=0

We add all the numbers together, and all the variables

36x^2-79=0

a = 36; b = 0; c = -79;
Δ = b2-4ac
Δ = 02-4·36·(-79)
Δ = 11376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{11376}=\sqrt{144*79}=\sqrt{144}*\sqrt{79}=12\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{79}}{2*36}=\frac{0-12\sqrt{79}}{72}
=-\frac{12\sqrt{79}}{72}
=-\frac{\sqrt{79}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{79}}{2*36}=\frac{0+12\sqrt{79}}{72}
=\frac{12\sqrt{79}}{72}
=\frac{\sqrt{79}}{6}
$