What is the solution set for this linear-quadratic system of equations?
y = x2 − x − 12
y − x − 3 = 0

Y - x - 3 = 0
y = x + 3

y = x^2 - x - 12
x + 3 = x^2 - x - 12
x^2 - x - 12 - x - 3 = 0
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0

x - 5 = 0     y = x + 3
x = 5          y = 5 + 3
                  y = 8

x + 3 = 0     y = x + 3
x = -3          y = -3 + 3
                   y = 0

so ur solutions are : (5,8) and (-3,0)