x^2+x+1=17

Solution for x^2+x+1=17 equation:

x^2+x+1=17

We move all terms to the left:

x^2+x+1-(17)=0

We add all the numbers together, and all the variables

x^2+x-16=0

a = 1; b = 1; c = -16;
Δ = b2-4ac
Δ = 12-4·1·(-16)
Δ = 65
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{-(1)-\sqrt{65}}{2*1}=\frac{-1-\sqrt{65}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*1}=\frac{-1+\sqrt{65}}{2}
$