I have this problem. Convert to standard form.
-2x^{2}-6x+10
The answer I come up with does not work when I check my work. What did I do wrong?
-2(x+1.5)^{2}+14.5
C-
x
It's hard to point out where you went wrong when we don't know what you actually did to get that answer...
$\displaystyle \begin{align*} -2 \, x^2 - 6 \, x + 10 &= -2 \, \left( x^2 + 3 \, x - 5 \right) \\ &= -2 \, \left[ x^2 + 3\,x + \left( \frac{3}{2} \right) ^2 - \left( \frac{3}{2} \right) ^2 - 5 \right] \\ &= -2 \, \left[ \left( x + \frac{3}{2} \right) ^2 - \frac{9}{4} - \frac{20}{4} \right] \\ &= -2 \, \left[ \left( x + \frac{3}{2} \right) ^2 - \frac{29}{4} \right] \\ &= -2 \, \left( x + \frac{3}{2} \right) ^2 + \frac{29}{2} \end{align*}$
I agree with your answer, why do you think it's wrong?
From my experience, most secondary algebra texts have standard form as $y=ax^2+bx+c$. The same texts usually name $y=a(x-h)^2+k$ as the vertex form.
I have seen the vertex form called standard form in other sources. I've also seen $y=ax^2+bx+c$ called the general form.
Depends on the text, I guess.