solving for the variable

[babyv]

n(5n+18)=8
I have moved the 8 over to the other side so now I have 5n^2+18n-8=0 which is (5n+_)(n-_) Im not even sure if the signs are right I know that the missing numbers are divisable by -8 that add to 18 but I can not find any numbers to fit...Does this mean there is no solution? :?:

 

[Mrspi]

n(5n+18)=8
I have moved the 8 over to the other side so now I have 5n^2+18n-8=0 which is (5n+_)(n-_) Im not even sure if the signs are right I know that the missing numbers are divisable by -8 that add to 18 but I can not find any numbers to fit...Does this mean there is no solution? :?:

Ok...so you have this:

5n[sup:3ks5wcrk]2[/sup:3ks5wcrk] + 18n - 8 = 0

You CAN, of course, use the quadratic formula to solve ANY quadratic equation, whether the solutions are real or not.

The solutions for

an^2 + bn + c = 0 are

n = [-b + sqrt(b[sup:3ks5wcrk]2[/sup:3ks5wcrk] - 4ac)] / (2a)

So, substituting the values for a, b, and c from your equation, you've got

n = [-18 + sqrt(18[sup:3ks5wcrk]2[/sup:3ks5wcrk] - 4*5*(-8)] / 2*5


Do the arithmetic:

n = [-18 + sqrt(324 + 160)] / 10

n = [-18 + sqrt(484) ] / 10

n = [-18 + 22] / 10

n = 4/10 OR n = -40/10

n = (2/5) OR n = -4