Hello from MrBillDoesMath!
Answer: 7/3
Discussion:
Using standard notation, the solutions of the quadratic ax^2 + bx + c = 0 are given by
x = ( -b +\- sqrt * b^2 - 4ac) ) / 2a
In our case a = 3, b = -19, and c = 28
x = ( -(-19) +\- sqrt ( (-19)^2 - 4*3*28) ) / ( 2*3)
= ( 19 +\- sqrt (361 - 336) ) / 6
= ( 19 +\- sqrt(25) ) /6
= ( 19 +\- 5) / 6
= (19 + 5) /6 = 24/6 = 4 and (19-5)/6 = 14/6 = 7/3
The other solution ( not equal to 4) is 7/3
Thank you,
MrB
