Your question is worded really well .. I can see you have tried it properly for yourself. I like that :)
7+- the square root of 9 divided by 2
I got it using the quadratic formula from x^2+4x-7=0
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-4 \pm \sqrt{4^2-4*1*-7} \over 2*1}\\ x = {-4 \pm \sqrt{16--28} \over 2}\\ x = {-4 \pm \sqrt{16+28} \over 2}\\ x = {-4 \pm \sqrt{44} \over 2}\\ x = {-4 \pm \sqrt{4*11} \over 2}\\ x = {-4 \pm2 \sqrt{11} \over 2}\\ x = {2(-2 \pm \sqrt{11} )\over 2}\\ x=-2 \pm \sqrt{11} \)
So if you did not mistype your original equation then you did something wrong with the quadratic formula.
Now I will look at you ACTUAL question ... which is simpler :)
7+- the square root of 9 divided by 2
\(x = {7 \pm \sqrt{9} \over 2}\\ x = {7 \pm 3 \over 2}\\ x = {7 + 3 \over 2} \qquad or \qquad x = {7 - 3 \over 2}\\ x = {10 \over 2} \qquad \;\; or \qquad x = {4 \over 2}\\ x = 5 \qquad \quad or\quad \quad x = 2\\\)
This would have been the solution to \(x^2-7x+10=0\)
.
so i'm going to get it until i get home
but i finally do
