Problem attached as picture below. (Algebra 1) Mai and Jada are solving the equation 2z-7=15different solutions.using the quadratic formula but foundMai wrote.74,/7-4(2)(-15)2(2)--7+,/49-(-120)69/16-7L134エー-5 0r z =Jada wrote:(-7)+/-7*-4(2)(-15)2(2)71,49-(-120)71 71a. If this equation is written in standard form, aI + br +e= 0what are the values of a,b, and c?Type the answers in the boxes below.b. Do you agree with either of them?Select the answer from the drop-down list.

Given equation is 2x2-7x=15 or. 2x2-7x-15=0

Mai and Jada are solving the equation 2r-71 = 15 using the quadratic formula but found different solutions. Mai wrote: -7+,/7-4(2)(-15) 2(2) -7+,/49-(-120) 4. -7+V169 4. -713 エー-5 or z= } Jada wrote: --7)+/-7-4(2)(-15) 2(2) 74/-49-(-120) 4. 74 71 a. If this equation is written in standard form, az + br + c=0 what are the values of a, b, and c? Type the answers in the boxes below. b. Do you agree with either of them?