Now solve `6x^{2}-17x+12=0`for x using the quadratic formula.
Remember to show your work.
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the equation is in the form ax^2 + bx + c = 0.
For the equation 6x^2 - 17x + 12 = 0, the coefficients are:
a = 6
b = -17
c = 12
Substituting these values into the quadratic formula:
x = (-(-17) ± √((-17)^2 - 4(6)(12))) / (2(6))
x = (17 ± √(289 - 288)) / 12
x = (17 ± √1) / 12
x = (17 ± 1) / 12
So, the two solutions are:
x = (17 + 1) / 12 = 18 / 12 = 3/2
x = (17 - 1) / 12 = 16 / 12 = 4/3
Therefore, the solutions to the equation 6x^2 - 17x + 12 = 0 are x = 3/2 and x = 4/3.