(2x+3)(2x-3)=4x^+5(2x-1)

Solution for (2x+3)(2x-3)=4x^+5(2x-1) equation:

(2x+3)(2x-3)=4x^+5(2x-1)

We move all terms to the left:

(2x+3)(2x-3)-(4x^+5(2x-1))=0

We use the square of the difference formula

4x^2-(4x^+5(2x-1))-9=0


We calculate terms in parentheses: -(4x^+5(2x-1)), so:

4x^+5(2x-1)

We add all the numbers together, and all the variables

4x+5(2x-1)

We multiply parentheses

4x+10x-5

We add all the numbers together, and all the variables

14x-5

Back to the equation:

-(14x-5)


We get rid of parentheses

4x^2-14x+5-9=0

We add all the numbers together, and all the variables

4x^2-14x-4=0

a = 4; b = -14; c = -4;
Δ = b2-4ac
Δ = -142-4·4·(-4)
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{65}}{2*4}=\frac{14-2\sqrt{65}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{65}}{2*4}=\frac{14+2\sqrt{65}}{8}
$