25x=2(x+4)(x+2)

Solution for 25x=2(x+4)(x+2) equation:

25x=2(x+4)(x+2)

We move all terms to the left:

25x-(2(x+4)(x+2))=0

We multiply parentheses ..

-(2(+x^2+2x+4x+8))+25x=0


We calculate terms in parentheses: -(2(+x^2+2x+4x+8)), so:

2(+x^2+2x+4x+8)

We multiply parentheses

2x^2+4x+8x+16

We add all the numbers together, and all the variables

2x^2+12x+16

Back to the equation:

-(2x^2+12x+16)


We add all the numbers together, and all the variables

25x-(2x^2+12x+16)=0

We get rid of parentheses

-2x^2+25x-12x-16=0

We add all the numbers together, and all the variables

-2x^2+13x-16=0

a = -2; b = 13; c = -16;
Δ = b2-4ac
Δ = 132-4·(-2)·(-16)
Δ = 41
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{-(13)-\sqrt{41}}{2*-2}=\frac{-13-\sqrt{41}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{41}}{2*-2}=\frac{-13+\sqrt{41}}{-4}
$