solve
15t^2+7t=2
15t^2 + 7t -2 = 0
(5t-1)(3t+2) = 0
t = -2/3, 1/5
t =(-b�}�ã(b^2-4ac))/2a
t =(-7�}�ã(7^2-4(15)(-2)))/(2(15))
t=(-7�}�ã(49 +120))/30
t=(-7�}�ã169)/30
t=(-7�}13)/30
t=(-7+13)/30 = 6/30 = 1/5
t=(-7-13)/30 = -(20/30) = -(2/3)
Therefore, t = 1/5 & -(2/3)
To solve the equation 15t^2 + 7t = 2, we can set the equation equal to zero by subtracting 2 from both sides:
15t^2 + 7t - 2 = 0
This is now a quadratic equation in standard form. To solve it, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 15, b = 7, and c = -2. Plugging these values into the formula, we have:
t = (-7 ± √(7^2 - 4 * 15 * -2)) / (2 * 15)
Simplifying the equation further:
t = (-7 ± √(49 + 120)) / 30
t = (-7 ± √169) / 30
t = (-7 ± 13) / 30
Now, we can find two solutions by evaluating both the positive and negative square root:
t1 = (-7 + 13) / 30 = 6 / 30 = 1/5
t2 = (-7 - 13) / 30 = -20 / 30 = -2/3
Therefore, the solutions to the equation 15t^2 + 7t = 2 are t = 1/5 and t = -2/3.