Answer:
Given 7b²-21b-273=7, the solutions are x1 = 8 and x2 = -5.
Step-by-step explanation:
Given 7b²-21b-273=7, first you need to equal zero. So
7b²-21b-273-7=0 ⇒ 7b²-21b-280 = 0
The secon step is to find the solutions applying Bhaskara´s formula x = (-b ± √(b²-4×a×c))/2×a
Where a=7, b= -21 and c= -280
After you identified each term, you have to replace it on the formula so....
x = (21 ± √(21² - 4×7×(-280)))/2×7 ⇒ x = (21 ± √(441 + 7840))/14 ⇒ x = (21 ± √8281)/14
Then you will obtain two values for x, called x1 = 8 and x2=-5.
7b^2 -21b -273 =7
Moving the -273 to the right hand side
7b^2 -21b =7+273
7b^2 -21b =280
Dividing the whole equation by 7,thus 7b^2/7 -21b/7 = 280/7
You will get
b^2 -3b =40.
Solving the equation as quadratic by moving the 40 to left side and equating the equation to zero, thus
b^2 -3b -40 =0
Factorizing b , you will get
(b-8) and (b+5) =0 thus b=8 ,b=-5 respectively. Now substituting the -5 in the equation satisfies it. Therefore b = -5
