25x2-9=(5x+3)(x-5)

Solution for 25x2-9=(5x+3)(x-5) equation:

25x^2-9=(5x+3)(x-5)

We move all terms to the left:

25x^2-9-((5x+3)(x-5))=0

We multiply parentheses ..

25x^2-((+5x^2-25x+3x-15))-9=0


We calculate terms in parentheses: -((+5x^2-25x+3x-15)), so:

(+5x^2-25x+3x-15)

We get rid of parentheses

5x^2-25x+3x-15

We add all the numbers together, and all the variables

5x^2-22x-15

Back to the equation:

-(5x^2-22x-15)


We get rid of parentheses

25x^2-5x^2+22x+15-9=0

We add all the numbers together, and all the variables

20x^2+22x+6=0

a = 20; b = 22; c = +6;
Δ = b2-4ac
Δ = 222-4·20·6
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{-(22)-2}{2*20}=\frac{-24}{40}
=-3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*20}=\frac{-20}{40}
=-1/2
$