5(x)=x^2+2x+1

Solution for 5(x)=x^2+2x+1 equation:

5(x)=x^2+2x+1

We move all terms to the left:

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

We get rid of parentheses

-x^2+5x-2x-1=0

We add all the numbers together, and all the variables

-1x^2+3x-1=0

a = -1; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·(-1)·(-1)
Δ = 5
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{-(3)-\sqrt{5}}{2*-1}=\frac{-3-\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5}}{2*-1}=\frac{-3+\sqrt{5}}{-2}
$