# Intermedia Algebra

Solve by factoring:
2a^2 − 3a = −5

1. 2a^2 - 3a = 5.

2a^2 - 3a - 5 = 0,
Use the AC Method:
A*C = 2*(-5) = 10,
2a^2 + (2a - 5a) -5 = 0,
Group the 4 terms into 2 factorable pairs:
(2a^2 + 2a) + (-5a - 5) = 0,
Factor each binomial:
2a(a + 1) -5(a + 1) = 0,
Factor out a+1:
(a + 1)(2a - 5) = 0,

a + 1 = 0,
a = -1.

2a - 5 = 0,
2a = 5,
a = 2.5.

Solution set: a = -1, and a = 2.5.

posted by Henry

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