# Thread: Solving this simple function

1. ## Solving this simple function

X^2-5x-3=0?

I honestly have no idea how to do this, perhaps x^2-5x=3, but then I'm still stuck, help!

2. Originally Posted by fattydq
X^2-5x-3=0?

I honestly have no idea how to do this, perhaps x^2-5x=3, but then I'm still stuck, help!
$x^2-5x-3=0$
It doesn't factorise easily, do you know the quadratic formula?
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ So in this case a=1 b=-5 and c=-3

3. leave the equation equal to 0 to solve.

since it can't be factorized you can solve either by

$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$

where a, b, c are the coefficients of the terms, or you can solve using 'completing the square'

$(x - \frac{5}{2})^2 - \frac{25}{4} - 3$

$(x - \frac{5}{2})^2 = \frac{37}{4}$

$x = \frac{5}{2} \pm \sqrt{\frac{37}{4}}$

4. Originally Posted by JeWiSh
$x^2-5x-3=0$
It doesn't factorise easily, do you know the quadratic formula?
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ So in this case a=1 b=-5 and c=-3
Oh yeah, I did already know that but for some reason my mind just blanked when I saw it. Thanks a bunch!