25=(2x-3)(2x+5)

Solution for 25=(2x-3)(2x+5) equation:

25=(2x-3)(2x+5)

We move all terms to the left:

25-((2x-3)(2x+5))=0

We multiply parentheses ..

-((+4x^2+10x-6x-15))+25=0


We calculate terms in parentheses: -((+4x^2+10x-6x-15)), so:

(+4x^2+10x-6x-15)

We get rid of parentheses

4x^2+10x-6x-15

We add all the numbers together, and all the variables

4x^2+4x-15

Back to the equation:

-(4x^2+4x-15)


We get rid of parentheses

-4x^2-4x+15+25=0

We add all the numbers together, and all the variables

-4x^2-4x+40=0

a = -4; b = -4; c = +40;
Δ = b2-4ac
Δ = -42-4·(-4)·40
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{41}}{2*-4}=\frac{4-4\sqrt{41}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{41}}{2*-4}=\frac{4+4\sqrt{41}}{-8}
$