(3x-4)(4x-3)=6x-7

Solution for (3x-4)(4x-3)=6x-7 equation:

(3x-4)(4x-3)=6x-7

We move all terms to the left:

(3x-4)(4x-3)-(6x-7)=0

We get rid of parentheses

(3x-4)(4x-3)-6x+7=0

We multiply parentheses ..

(+12x^2-9x-16x+12)-6x+7=0

We get rid of parentheses

12x^2-9x-16x-6x+12+7=0

We add all the numbers together, and all the variables

12x^2-31x+19=0

a = 12; b = -31; c = +19;
Δ = b2-4ac
Δ = -312-4·12·19
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-7}{2*12}=\frac{24}{24}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+7}{2*12}=\frac{38}{24}
=1+7/12
$