The quadratic equation 3x^2 -19x + 28 = 0 has a solution of x = 4. what is the other solution? * free response question

Hello from MrBillDoesMath!

    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

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer