c^2-3c=28

Solution for c^2-3c=28 equation:

c^2-3c=28

We move all terms to the left:

c^2-3c-(28)=0

a = 1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*1}=\frac{-8}{2}
=-4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*1}=\frac{14}{2}
=7
$