can someone help me out factor out this
0=16t^2-122t+180
i have so far this but the last one i am not sure
8(2t^2-14t + ?)
you need to use the quadratic formula: -b(+or-)[sqrt(b^2-4ac)/2a]
so you are saying that i don't have to factor it out any further i can just continue to do the quadratic formula
yes.
Unless the coefficient of the square term divides into the coefficient of the middle term leaving an even answer, the method of completing the square to solve a quadratic is a poor choice.
Use the quadratic formula.
To factorize the given quadratic equation 0=16t^2-122t+180, we can use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions can be found using the following formula:
x = (-b +/- sqrt(b^2 - 4ac)) / (2a)
In this case, the equation is in the form of 16t^2 - 122t + 180 = 0. So we can directly apply the quadratic formula.
First, let's identify the values of a, b, and c:
a = 16
b = -122
c = 180
Next, substitute these values into the quadratic formula:
t = (-(-122) +/- sqrt((-122)^2 - 4 * 16 * 180)) / (2 * 16)
Now let's simplify this expression:
t = (122 +/- sqrt(14884 - 11520)) / 32
t = (122 +/- sqrt(3356)) / 32
t = (122 +/- sqrt(4 * 839)) / 32
t = (122 +/- 2sqrt(839)) / 32
t = (61 +/- sqrt(839)) / 16
So, the solutions to the quadratic equation 0=16t^2-122t+180 are t = (61 + sqrt(839))/16 and t = (61 - sqrt(839))/16.
Therefore, it is not possible to factorize the quadratic equation any further. The quadratic formula provides us with the exact solutions for the equation.