What are the roots of the polynomial equation x^2+8x-48=0

 Given polynomial x^2+8x-48 = 0

x^2+12x-4x-48 = 0

x(x+12)-4(x+12) = 0

(x+12)(x-4) = 0

x+12 = 0

Subtract 12 from each side.

x+12-12 = 0-12

x = -12

and x-4 = 0

Add 4 to each side.

x-4+4 = 0+4

x = 4

Roots are -12,4.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)