(2x-3)(x=4)=6

Solution for (2x-3)(x=4)=6 equation:

(2x-3)(x=4)=6

We move all terms to the left:

(2x-3)(x-(4))=0

We multiply parentheses ..

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

We get rid of parentheses

2x^2-8x-3x+12=0

We add all the numbers together, and all the variables

2x^2-11x+12=0

a = 2; b = -11; c = +12;
Δ = b2-4ac
Δ = -112-4·2·12
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*2}=\frac{6}{4}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*2}=\frac{16}{4}
=4
$