(2x+1)(4x-3)=3(4x-3)

Solution for (2x+1)(4x-3)=3(4x-3) equation:

(2x+1)(4x-3)=3(4x-3)

We move all terms to the left:

(2x+1)(4x-3)-(3(4x-3))=0

We multiply parentheses ..

(+8x^2-6x+4x-3)-(3(4x-3))=0


We calculate terms in parentheses: -(3(4x-3)), so:

3(4x-3)

We multiply parentheses

12x-9

Back to the equation:

-(12x-9)


We get rid of parentheses

8x^2-6x+4x-12x-3+9=0

We add all the numbers together, and all the variables

8x^2-14x+6=0

a = 8; b = -14; c = +6;
Δ = b2-4ac
Δ = -142-4·8·6
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{-(-14)-2}{2*8}=\frac{12}{16}
=3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2}{2*8}=\frac{16}{16}
=1
$