I have the folowing equation
$$\frac{16}{3}x^2 - 2x -45 = 0$$ I am trying to solve it for about 40 minutes. I am starting to think I did something wrong. Maybe the equation insn't right, I derived it.
I got as far as this
(16/3)x2 - 2x -45 = 0 (-45 the signs will be different)
1 . 45
16/3 1 3 . 15
4/3 4 9 . 5
2/3 8
The pairs below the equation are the products. I cannot find the pairs.
$$\frac{16}3x^2 - 2x -45 = 0\\ 16x^2 - 6x -135 = 0\\ \left(4x-\frac34\right)^2 -\frac{3^2}{4^2} -135 = 0\\ \left(4x-\frac34\right)^2 = 135 + \frac{9}{16} = \frac{2169}{16}$$ You can finish the rest yourself. The right hand side is not a nice square, so the solutions are not rational numbers, and it is therefore not surprising that they cannot be found by trying to factorise the original quadratic. Directly applying the quadratic formula is easier than completing the square in a case like this.