(2x+1)2=4(x2-1)+x-1

Solution for (2x+1)2=4(x2-1)+x-1 equation:

(2x+1)2=4(x2-1)+x-1

We move all terms to the left:

(2x+1)2-(4(x2-1)+x-1)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

-(4(+x^2-1)+x-1)+4x+2=0


We calculate terms in parentheses: -(4(+x^2-1)+x-1), so:

4(+x^2-1)+x-1

We multiply parentheses

4x^2+x-4-1

We add all the numbers together, and all the variables

4x^2+x-5

Back to the equation:

-(4x^2+x-5)


We add all the numbers together, and all the variables

4x-(4x^2+x-5)+2=0

We get rid of parentheses

-4x^2+4x-x+5+2=0

We add all the numbers together, and all the variables

-4x^2+3x+7=0

a = -4; b = 3; c = +7;
Δ = b2-4ac
Δ = 32-4·(-4)·7
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*-4}=\frac{-14}{-8}
=1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*-4}=\frac{8}{-8}
=-1
$