(3x+5)(3x-5)-(6-4x)(6+4x)=0

Solution for (3x+5)(3x-5)-(6-4x)(6+4x)=0 equation:

(3x+5)(3x-5)-(6-4x)(6+4x)=0

We add all the numbers together, and all the variables

(3x+5)(3x-5)-(-4x+6)(4x+6)=0

We use the square of the difference formula

9x^2-(-4x+6)(4x+6)-25=0

We multiply parentheses ..

9x^2-(-16x^2-24x+24x+36)-25=0

We get rid of parentheses

9x^2+16x^2+24x-24x-36-25=0

We add all the numbers together, and all the variables

25x^2-61=0

a = 25; b = 0; c = -61;
Δ = b2-4ac
Δ = 02-4·25·(-61)
Δ = 6100
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:

$\sqrt{\Delta}=\sqrt{6100}=\sqrt{100*61}=\sqrt{100}*\sqrt{61}=10\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{61}}{2*25}=\frac{0-10\sqrt{61}}{50}
=-\frac{10\sqrt{61}}{50}
=-\frac{\sqrt{61}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{61}}{2*25}=\frac{0+10\sqrt{61}}{50}
=\frac{10\sqrt{61}}{50}
=\frac{\sqrt{61}}{5}
$