How do you solve #a^2 + 2a - 48 = 0#?

1 Answer
Feb 26, 2016

Answer:

This trinomial is factorable, so we can solve by factoring.

Explanation:

To factor a trinomial of the form #y = ax^2 + bx + c, a = 1#, we must find two numbers that multiply to c and that add to b.

Two numbers that multiply to -48 and that add to 2 are +8 and -6.

#(a + 8)(a - 6) = 0#

#a = -8 and 6#

Hopefully this helps.