x2-4x(2x-3)=5

Solution for x2-4x(2x-3)=5 equation:

x2-4x(2x-3)=5

We move all terms to the left:

x2-4x(2x-3)-(5)=0

We add all the numbers together, and all the variables

x^2-4x(2x-3)-5=0

We multiply parentheses

x^2-8x^2+12x-5=0

We add all the numbers together, and all the variables

-7x^2+12x-5=0

a = -7; b = 12; c = -5;
Δ = b2-4ac
Δ = 122-4·(-7)·(-5)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*-7}=\frac{-14}{-14}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*-7}=\frac{-10}{-14}
=5/7
$