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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-11x+5=0

a = 2; b = -11; c = +5;
Δ = b2-4ac
Δ = -112-4·2·5
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-9}{2*2}=\frac{2}{4}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+9}{2*2}=\frac{20}{4}
=5
$