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

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

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

We move all terms to the left:

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

We use the square of the difference formula

4x^2+(x+3)(4x-5)+9=0

We multiply parentheses ..

4x^2+(+4x^2-5x+12x-15)+9=0

We get rid of parentheses

4x^2+4x^2-5x+12x-15+9=0

We add all the numbers together, and all the variables

8x^2+7x-6=0

a = 8; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·8·(-6)
Δ = 241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{241}}{2*8}=\frac{-7-\sqrt{241}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{241}}{2*8}=\frac{-7+\sqrt{241}}{16}
$