Question

t^2+t+0.75 find zeroes of this quadratic equation​

Answer 1

Step-by-step explanation:

Example 1 :

2x2 - 5x + 2 = 0

Solution :

Comparing 2x2 - 5x + 2 = 0 and ax2 + bx + c = 0, we get

a = 2, b = -5 and c = 2

Then,

x = [-b ± √b2 - 4ac] / 2a

x = [-(-5) ± √(-5)2 - 4(2)(2)] / 2(2)

x = [5 ± √(25 - 16)] / 4

x = [5 ± √9] / 4

x = [5 ± 3] / 4

x = (5 + 3) /4 and x = (5 - 3)/4

x = 8/4 and x = 2/4

x = 2 and x = 1/2

Therefore, the solution is {1/2, 2}.

Answer 2

Answer:

1x+1−1×−2=−1

Multiply 1 and −2 to get −2.

1x+1−(−2)=−1

The opposite of −2 is 2.

1x+1+2=−1

Add 1 and 2 to get 3.

1x+3=−1

Subtract 3 from both sides.

1x=−1−3

Subtract 3 from −1 to get −4.

1x=−4

Reorder the terms.

x=−4