c^2+c=23

Solution for c^2+c=23 equation:

c^2+c=23

We move all terms to the left:

c^2+c-(23)=0

a = 1; b = 1; c = -23;
Δ = b2-4ac
Δ = 12-4·1·(-23)
Δ = 93
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{93}}{2*1}=\frac{-1-\sqrt{93}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{93}}{2*1}=\frac{-1+\sqrt{93}}{2}
$