Perhaps you could revisit the quadratic roots solution ...
for f(x) = ax^2 + bx + c
the roots are
x1 = (- b + (b^2 - 4ac)^0.5)/2a
x2 = (- b - (b^2 - 4ac)^0.5)/2a
I presume that the 15x is a typo and you meant to write
2x^2 - 7x - 15 = 0
which gives roots (assuming that I have counted all my brackets correctly and put in the correct numbers ...)
x = (-(-7) +/- ((-7)^2 - 4(2)(-15))^0.5)/2(2)
= (7 +/- 13)/4
= 20/4 and -6/4
= 5 and -1.5
If what you wrote is correct and what was intended then the equation is a quadratic whose constant term is zero and the solution comes down to zero and 11.
Last edited by john_tullamarine; 6th January 2003 at 22:42.