“The points of intersection of the graph of $y=3+x-0.5x^2$ and the line $y=k$ are the solutions of the equation $10+2x-x^2=0$”

I was thinking that maybe i could find the solutions for the second equation and then find value of $k$. But thats too much work for one mark, and anyway it says to use the line $y=k$ to find the solutions. So i am supposed to find k through another method. Please help. I have my cambridge maths exam on 30 may

• Please consider using mathjax – Digitalis May 26 '18 at 8:53
• But what is the question in the problem you mention? – Bernard May 26 '18 at 8:54
• Good luck for your exam! Also k=-2 – Tony Hellmuth May 26 '18 at 8:55
• The first senctence is completely wrong – Fakemistake May 26 '18 at 8:58

$$-x^2+2x+10=0\iff -\frac12 x^2+x+3=-2$$
then $k=-2$.
$$10+2x-x^2=0$$ $$x=1-\sqrt{11} \text{ or }x=1+\sqrt{11}$$ $$y=3+x−0.5x^2$$ $$y=3+(1-\sqrt{11})−0.5(1-\sqrt{11})^2$$ $$y=-2$$ $$y=3+(1+\sqrt{11})−0.5(1+\sqrt{11})^2$$ $$y=-2$$
Therefore since both points lie on $y=-2$, $k=-$2