Originally Posted by
gregzaj1 Positive root of the following equation:
34*x^2 + 68*x - 510
First, you understand that this is NOT an equation, don't you? I think you meant $\displaystyle 34x^2+ 68x- 510= 0$ which is an equation because it has "=" but I had to guess at the "0"
Again, that is NOT an equation. You mean $\displaystyle ax^2+ bx+ c= 0$.
Now, compare that with your $\displaystyle 34x^2+ 68x- 510= 0$. You should see that a= 34, b= 68, and c= -510
x1 = ( - b + sqrt ( b*b - 4*a*c ) ) / ( 2*a)
So "b*b- 4*a*c" is (68)(68)- 4(34)(-510)= 4624+ 69360= 73984 and the square root is 272.
-b+sqrt(b*b- 4*a*c)= -68+ 272= 204. 2*a= 2*34= 68 so x= 204/68= 3.
That wasn't so hard was it?
(If you are very clever, like richard1234, you might recognise that 68= 2(34) and 510= 15(34) so dividing through by 34 at the start simplifies the numbers.)
I have no idea how to go about doing this. Can someone help me out?