Question

How many solutions does each equation have?
6x^2+13x-5=0 and 9x^2+6x+1=0

Answer 1

The number of solutions is equal to the highest exponent. Which in this case is 2 for both of them.

Answer 2

can you show the work on it?

Answer 3

\[-b \pm \sqrt{b ^{2} - 4ac}/2a\] is the quadratic formula. For the first problem \[6x^2 +13x -5\] a = 6 b = 13 and c = -5 so plugging these numbers into the quadratic formula you get \[-13 \pm \sqrt{13^2 - 4(6)(-5)}/12\] simplifying the expression in the radical you get \[-13 \pm \sqrt{289}/12\] so one solution is \[-13 + \sqrt{289}/12\] and the other is \\[-13 - \sqrt{289}/12\] The whole thing is divided by 12 not just the radical just to make that clear. You do the same thing for the other problem