Question

Factorise
n²+241 n-1332=0​

Answer 1

On comparing n²+241 n-1332=0 with the standard equation a$x^{2}$ + bx + c

⇒ a=1 , b= 241 , c=-1332

Solving    n2+241n-1332 = 0 by the Quadratic Formula .

According to the Quadratic Formula,  n  , the solution for   An2+Bn+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                   

           - B  ±  √ B2-4AC

 n =   ————————

                     2A

 In our case,  A   =     1

                     B   =   241

                     C   =  -1332

Accordingly,  B2  -  4AC   =

                    58081 - (-5328) =

                    63409

Applying the quadratic formula :

              -241 ± √ 63409

  n  =    ————————

                       2

 √ 63409   , rounded to 4 decimal digits, is  251.8114

So now we are looking at:

          n  =  ( -241 ±  251.811 ) / 2

Two real solutions:

n =(-241+√63409)/2= 5.406

or:

n =(-241-√63409)/2=-246.406 Answer: