Solve the equation in the image.

b= 3, -6

Step-by-step explanation:

square both sides

b²+3b-2 = 4² = 16

b²+3b-2-16 =0

b²+3b-18 =0

b²+6b-3b-18=0

b(b+6)-3(b+6) =0

(b+6)(b-3)=0

b+6=0    b-3 =0

b= -6      b=3

b= 3, -6

Step-by-step explanation:

The given equation is

√b² + 3b - 2 = 4

The first step is to square the left hand side and the right hand side of the equation. It becomes

(√b² + 3b - 2)² = 4²

b² + 3b - 2 = 16

b² + 3b - 2 - 16 = 0

b² + 3b - 18 = 0

We would find two numbers such that their sum or difference is 3b and their product is -18b². The two numbers are 6b and - 3b. Therefore

b² + 6b - 3b - 18 = 0

b(b + 6) - 3(b + 6) = 0

b - 3 = 0 or b + 6 = 0

b = 3 or b = - 6

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

i believe the correct answer is b.