10(x^2-1)=1+10(x-1)

Solution for 10(x^2-1)=1+10(x-1) equation:

10(x^2-1)=1+10(x-1)

We move all terms to the left:

10(x^2-1)-(1+10(x-1))=0

We multiply parentheses

10x^2-(1+10(x-1))-10=0


We calculate terms in parentheses: -(1+10(x-1)), so:

1+10(x-1)

determiningTheFunctionDomain
10(x-1)+1

We multiply parentheses

10x-10+1

We add all the numbers together, and all the variables

10x-9

Back to the equation:

-(10x-9)


We get rid of parentheses

10x^2-10x+9-10=0

We add all the numbers together, and all the variables

10x^2-10x-1=0

a = 10; b = -10; c = -1;
Δ = b2-4ac
Δ = -102-4·10·(-1)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{35}}{2*10}=\frac{10-2\sqrt{35}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{35}}{2*10}=\frac{10+2\sqrt{35}}{20}
$