Michael B. answered • 02/10/13

Seasoned and experienced tutor with extensive science background

I'm going to assume that this was entered in wrong and that you meant:

4t^{2}-13t-12 = 0

This is a quadratic equation of the form:

ax^{2}+bx+c = 0

The factored solution in this case would be:

(a_{1}t+c_{1})(a_{2}t+c_{2})

a = 4 and is the product (a_{1}*a_{2}) so that the factors a_{1} and a_{2} must be (1*4) or (2*2) and are both positive

c = -12 and is the product (c_{1}*c_{2}) so that the factors c_{1} and c_{2} must be

(1*12), (2*6), or (3*4) where one factor is negative and the other is positive

b = -13 and is a sum of products meaning that the factors of a and c have to be cross multiplied and add up to -13:

(a_{1}*c_{2})+(a_{2}*c_{1}) = -13

The only case for b = -13 has 4*-4 = -16 and 1*3 = 3

Your answer is: (4t+3)(t-4)