7b squared - 21b-273=7

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

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.