<s>[INST] solve quadratic equation I'm trying to solve the following equation $2t^2 + t - 3 = 0$
I start by dividing by 2, $t^2 + \frac {t}{2} - \frac {3}{2} = 0$
Then I solve for t $t = - \frac{ \frac {1}{2} }{2} \binom{+}{-} \sqrt{(\frac {1}{2})^2 + \frac {3}{2}}$
$t = - \frac{1}{4} \binom{+}{-} \sqrt{(\frac {1}{4}) + \frac {6}{4}}$
I calculate $t = - \frac{1}{4} \binom{+}{-} \frac {\sqrt7}{4}$
$t_1 = - \frac{1}{4} + \frac {\sqrt7}{4}$
$t_2 = - \frac{1}{4} - \frac {\sqrt7}{4}$
But according to wolframalpha it's suppose to be 
$t_1 = 1$
$t_2 = - \frac {3}{2}$
Can't figure out where did I go wrong in my calculation?
 [/INST] Well actually you have applied the quadratic formula wrong. The roots of the equation $$ax^2+bx+c=0$$ is given by
$$\alpha, \beta ={-b \pm \sqrt {b^2-4ac} \over 2a}$$
So for the equation $$t^2+ \frac12t - \frac32=0$$
the roots become $$\alpha, \beta={- \frac12 \pm \sqrt { (\frac12)^2+4(\frac32)} \over 2}$$
which gives the roots as $1, -1.5$
But in cases like these, I'd suggest an alternative method.
the equation can be re-written as $$2t^2-2t+3t-3=0$$
which can be written as
$$2t(t-1)+3(t-1)=0$$
Now on taking $t-1$ as the common factor, we can write $$(t-1)(2t+3)=0$$
which gives the desired roots.
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