(2x-10)(3x-20)=1

Solution for (2x-10)(3x-20)=1 equation:

(2x-10)(3x-20)=1

We move all terms to the left:

(2x-10)(3x-20)-(1)=0

We multiply parentheses ..

(+6x^2-40x-30x+200)-1=0

We get rid of parentheses

6x^2-40x-30x+200-1=0

We add all the numbers together, and all the variables

6x^2-70x+199=0

a = 6; b = -70; c = +199;
Δ = b2-4ac
Δ = -702-4·6·199
Δ = 124
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:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-2\sqrt{31}}{2*6}=\frac{70-2\sqrt{31}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+2\sqrt{31}}{2*6}=\frac{70+2\sqrt{31}}{12}
$