Tx²+3x-7=0 has two real solution. what can be the deducted about the value of t?

value of t is greater than equal to - 9 / 28.

Step-by-step explanation:

Given Quadratic Polynomial : tx² + 3x - 7 = 0

Also, It has real solutions.

Standard Quadratic equation, is ax² + bx + c = 0

here, Determinant, D = b² - 4ac

decides nature of the roots.

if D < 0, roots / solutions are complex

if D = 0, roots are real and equal.

if D > 0, roots are real and different.

As given roots are real solutions.

Means Dis either equal to 0 or greater than 0

when D = 0

we have, 3² - 4 * t * (-7) = 0

9 + 28t = 0

t = - 9 / 28

when D > 0

we have, 3² - 4 * t * (-7) > 0

9 + 28t > 0

t > - 9 / 28

Therefore, value of t is greater than equal to - 9 / 28.