How this 0=t2−t−2 become this 0=(t−2)(t+1)

In order to t² - t -2 = 0 become into this (t - 2)(t + 1) = 0, we have to know that a  polynomial of the form ax² + bx + c = 0 can be factored as a(x + r1)(x + r2) = 0, where r1 and r2 are the roots of the equation ax² + bx + c = 0 and we can find them using the Quadratic Formula.

Let's solve the equation step-by-step.

t² - t -2 = 0 has the form at² + bt + c = 0, so can be factored as a(t + r1)(t + r2) = 0. With a = 1, b = -1, and c = -2

Step 1. Use the Quadratic Formula to find the roots r1 and r2.

t =  −b ± √b²−4ac/2a

t = -(-1) ± √(-1)²−4(1)(-2)/2(1)

t = 1 ± √1+8/2

t = 1 ± √9/2 where r1 = 1 + √9/2 and r2 = 1 - √9/2

Step 2. Calculate the roots r1 and r2

r1 = 1 + √9/2 = 2

r2 = 1 - √9/2 = -1

Step 3. Write the factored equation a(t + r1)(t + r2) = 0, with a = 1 and the values of r1 and r2 with opposite signs.

1(t - 2)(t + 1) = 0

(t - 2)(t + 1) = 0

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.