25(x-5)=(x+19)(x-5)

Solution for 25(x-5)=(x+19)(x-5) equation:

25(x-5)=(x+19)(x-5)

We move all terms to the left:

25(x-5)-((x+19)(x-5))=0

We multiply parentheses

25x-((x+19)(x-5))-125=0

We multiply parentheses ..

-((+x^2-5x+19x-95))+25x-125=0


We calculate terms in parentheses: -((+x^2-5x+19x-95)), so:

(+x^2-5x+19x-95)

We get rid of parentheses

x^2-5x+19x-95

We add all the numbers together, and all the variables

x^2+14x-95

Back to the equation:

-(x^2+14x-95)


We add all the numbers together, and all the variables

25x-(x^2+14x-95)-125=0

We get rid of parentheses

-x^2+25x-14x+95-125=0

We add all the numbers together, and all the variables

-1x^2+11x-30=0

a = -1; b = 11; c = -30;
Δ = b2-4ac
Δ = 112-4·(-1)·(-30)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*-1}=\frac{-12}{-2}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*-1}=\frac{-10}{-2}
=+5
$