Solve by factoring:

10t^2 – t − 2 = 0

10t^2 - t - 2 = 0.

Use the AC Method:
A*C = 10*(-2) = -20 = -5*4.
Select the pair of factors whose sum = -1:
10t^2 - 5t+4t - 2 = 0,
Form 2 factorable pairs by grouping:
(10t^2-5t) + (4t-2) = 0,
Factor each binomial:
5t(27-1) + 2(2x-1) = 0,
(2t-1)(5t+2) = 0,

2t-1 = 0,
2t = 1,
t = 1/2.

5t+2 = 0,
5t = -2,
t = -2/5.

Solution Set: t = 1/2, and-2/5.

Correction:

5t(27-1) should be 5t(2t-1).

All variables should be t and NOT x.

To solve the quadratic equation 10t^2 – t – 2 = 0 by factoring, we need to find two binomials whose product is equal to the given quadratic equation.

Step 1: Multiply the coefficient of the quadratic term (10) with the constant term (-2). This gives us -20.

Step 2: We need to find two numbers that multiply to give -20 and add up to the coefficient of the linear term (-1). The numbers -5 and 4 satisfy this condition since (-5) * 4 = -20 and (-5) + 4 = -1.

Step 3: Rewrite the middle term (-t) using the two numbers (-5 and 4) as coefficients. This gives us 10t^2 - 5t + 4t - 2 = 0.

Step 4: Group the terms and factor by grouping. We group the first two terms and the last two terms:
(10t^2 - 5t) + (4t - 2) = 0
5t(2t - 1) + 2(2t - 1) = 0

Step 5: Notice that the terms (2t - 1) appear in both groups. Factor out (2t - 1):
(2t - 1)(5t + 2) = 0

Step 6: Set each factor equal to zero and solve for t:
2t - 1 = 0 or 5t + 2 = 0

Step 7: Solve the equations:
2t = 1 or 5t = -2
t = 1/2 or t = -2/5

Therefore, the solutions to the quadratic equation 10t^2 – t – 2 = 0 are t = 1/2 and t = -2/5.