(6x-10)(2x-1)=40

Solution for (6x-10)(2x-1)=40 equation:

(6x-10)(2x-1)=40

We move all terms to the left:

(6x-10)(2x-1)-(40)=0

We multiply parentheses ..

(+12x^2-6x-20x+10)-40=0

We get rid of parentheses

12x^2-6x-20x+10-40=0

We add all the numbers together, and all the variables

12x^2-26x-30=0

a = 12; b = -26; c = -30;
Δ = b2-4ac
Δ = -262-4·12·(-30)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-46}{2*12}=\frac{-20}{24}
=-5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+46}{2*12}=\frac{72}{24}
=3
$