1/2x^2+3/4x-1=0
-3/4(+-)SQRT((3/4^2)-4(1/2)(-1) / 2(1/2)
-3/4(+-)SQRT(9/16+2) / 1
This is as far as I can get I do not see any errors yet the answer is supposed to be
-3+-SQRT(41) /4
I am clueless as to where the 41 comes from
1/2x^2+3/4x-1=0
-3/4(+-)SQRT((3/4^2)-4(1/2)(-1) / 2(1/2)
-3/4(+-)SQRT(9/16+2) / 1
This is as far as I can get I do not see any errors yet the answer is supposed to be
-3+-SQRT(41) /4
I am clueless as to where the 41 comes from
The original problem:
$\displaystyle \dfrac{1}{2}\,x^{2}+\dfrac{3}{4}\,x-1=0.$
Quadratic formula gives
$\displaystyle x=\dfrac{-3/4\pm\sqrt{9/16-4(1/2)(-1)}}{1}=-3/4\pm\sqrt{9/16+2}=
-3/4\pm\sqrt{(9+32)/16}=\dfrac{-3\pm\sqrt{41}}{4}.$
That's what I get. WolframAlpha gives the same.
I conclude your book's answer is wrong.
Obviously it isn't necessary, but it would be easier to calculate if you multiplied the equation by four.
$\displaystyle 2x^2 + 3x - 4 = 0$ is so much easier to substitute into the formula, and it will probably save time as it's so much simpler to follow through.
To start, that should be: [-3 +- SQRT(41)] /4 ; the extra brackets are IMPORTANT.
GOOD IDEA to always get rid of fractions before doing anything else;
so multiply your equation by 4: 2x^2 + 3x - 4 = 0 : get the picture?
Now use the quadratic and you'll have no problems