(3x-5)(4x-5)=28

Solution for (3x-5)(4x-5)=28 equation:

(3x-5)(4x-5)=28

We move all terms to the left:

(3x-5)(4x-5)-(28)=0

We multiply parentheses ..

(+12x^2-15x-20x+25)-28=0

We get rid of parentheses

12x^2-15x-20x+25-28=0

We add all the numbers together, and all the variables

12x^2-35x-3=0

a = 12; b = -35; c = -3;
Δ = b2-4ac
Δ = -352-4·12·(-3)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-37}{2*12}=\frac{-2}{24}
=-1/12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+37}{2*12}=\frac{72}{24}
=3
$