7(2x-1)(x-4)-3(3x-1)(x-4)=21

Solution for 7(2x-1)(x-4)-3(3x-1)(x-4)=21 equation:

7(2x-1)(x-4)-3(3x-1)(x-4)=21

We move all terms to the left:

7(2x-1)(x-4)-3(3x-1)(x-4)-(21)=0

We multiply parentheses ..

7(+2x^2-8x-1x+4)-3(3x-1)(x-4)-21=0

We multiply parentheses

14x^2-56x-7x-3(3x-1)(x-4)+28-21=0

We multiply parentheses ..

14x^2-3(+3x^2-12x-1x+4)-56x-7x+28-21=0

We add all the numbers together, and all the variables

14x^2-3(+3x^2-12x-1x+4)-63x+7=0

We multiply parentheses

14x^2-9x^2+36x+3x-63x-12+7=0

We add all the numbers together, and all the variables

5x^2-24x-5=0

a = 5; b = -24; c = -5;
Δ = b2-4ac
Δ = -242-4·5·(-5)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-26}{2*5}=\frac{-2}{10}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+26}{2*5}=\frac{50}{10}
=5
$