x=x^-49-(5x+3)(x+7)

Solution for x=x^-49-(5x+3)(x+7) equation:

x=x^-49-(5x+3)(x+7)

We move all terms to the left:

x-(x^-49-(5x+3)(x+7))=0

We multiply parentheses ..

-(x^-49-(+5x^2+35x+3x+21))+x=0


We calculate terms in parentheses: -(x^-49-(+5x^2+35x+3x+21)), so:

x^-49-(+5x^2+35x+3x+21)

determiningTheFunctionDomain
-(+5x^2+35x+3x+21)+x^-49

We add all the numbers together, and all the variables

-(+5x^2+35x+3x+21)+x-49

We get rid of parentheses

-5x^2-35x-3x+x-21-49

We add all the numbers together, and all the variables

-5x^2-37x-70

Back to the equation:

-(-5x^2-37x-70)


We get rid of parentheses

5x^2+37x+x+70=0

We add all the numbers together, and all the variables

5x^2+38x+70=0

a = 5; b = 38; c = +70;
Δ = b2-4ac
Δ = 382-4·5·70
Δ = 44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{11}}{2*5}=\frac{-38-2\sqrt{11}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{11}}{2*5}=\frac{-38+2\sqrt{11}}{10}
$