Find the center of the circle that can be circumscribed about efg with

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629

12r - 8 - 12

you have to combine like terms. below i have bolded the like terms:

12r - 8 - 12

together it combines to:

12r - 20

hope this !

c(x)= 6x + 3

step-by-step explanation: