25x^2+20x+1=(x+2)(x+11)

Solution for 25x^2+20x+1=(x+2)(x+11) equation:

25x^2+20x+1=(x+2)(x+11)

We move all terms to the left:

25x^2+20x+1-((x+2)(x+11))=0

We multiply parentheses ..

25x^2-((+x^2+11x+2x+22))+20x+1=0


We calculate terms in parentheses: -((+x^2+11x+2x+22)), so:

(+x^2+11x+2x+22)

We get rid of parentheses

x^2+11x+2x+22

We add all the numbers together, and all the variables

x^2+13x+22

Back to the equation:

-(x^2+13x+22)


We add all the numbers together, and all the variables

25x^2+20x-(x^2+13x+22)+1=0

We get rid of parentheses

25x^2-x^2+20x-13x-22+1=0

We add all the numbers together, and all the variables

24x^2+7x-21=0

a = 24; b = 7; c = -21;
Δ = b2-4ac
Δ = 72-4·24·(-21)
Δ = 2065
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{-(7)-\sqrt{2065}}{2*24}=\frac{-7-\sqrt{2065}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{2065}}{2*24}=\frac{-7+\sqrt{2065}}{48}
$