(x-4)(8x-3)=x2-7x+12

Solution for (x-4)(8x-3)=x2-7x+12 equation:

(x-4)(8x-3)=x2-7x+12

We move all terms to the left:

(x-4)(8x-3)-(x2-7x+12)=0

We add all the numbers together, and all the variables

-(+x^2-7x+12)+(x-4)(8x-3)=0

We get rid of parentheses

-x^2+7x+(x-4)(8x-3)-12=0

We multiply parentheses ..

-x^2+(+8x^2-3x-32x+12)+7x-12=0

We add all the numbers together, and all the variables

-1x^2+(+8x^2-3x-32x+12)+7x-12=0

We get rid of parentheses

-1x^2+8x^2-3x-32x+7x+12-12=0

We add all the numbers together, and all the variables

7x^2-28x=0

a = 7; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·7·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*7}=\frac{0}{14}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*7}=\frac{56}{14}
=4
$