(1=2x+x^2)-(50+25x+2x^2)=

Solution for (1=2x+x^2)-(50+25x+2x^2)= equation:

(1=2x+x^2)-(50+25x+2x^2)=

We move all terms to the left:

(1-(2x+x^2)-(50+25x+2x^2))=0


We calculate terms in parentheses: +(1-(2x+x^2)-(50+25x+2x^2)), so:

1-(2x+x^2)-(50+25x+2x^2)

determiningTheFunctionDomain
-(2x+x^2)-(50+25x+2x^2)+1

We get rid of parentheses

-x^2-2x^2-2x-25x-50+1

We add all the numbers together, and all the variables

-3x^2-27x-49

Back to the equation:

+(-3x^2-27x-49)


We get rid of parentheses

-3x^2-27x-49=0

a = -3; b = -27; c = -49;
Δ = b2-4ac
Δ = -272-4·(-3)·(-49)
Δ = 141
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{-(-27)-\sqrt{141}}{2*-3}=\frac{27-\sqrt{141}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{141}}{2*-3}=\frac{27+\sqrt{141}}{-6}
$