3x^2-4x+8=8x^2-5x-3
\(3x^2-4x+8=8x^2-5x-3\\ 4x^2-x-11=0\) 8 - 3 = 5 not 4 !!
a b c
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = {1 \pm \sqrt{1-4\cdot 4\cdot (-11)} \over 2\cdot 4}\)
\(x=\frac{1\pm \sqrt{177}}{8}=0.125\pm 1.6630168\)
\(x_1=1.7880168\\ x_2=-1.5380168\)
This result is correct for (\(4x^2-x-11=0\)),
but not for (\(3x^2-4x+8=8x^2-5x-3\)).
Why ?
8 - 3 = 5 not 4 !!
Thanks for the help Alan and heureka!
I will correct immediately.
\(3x^2-4x+8=8x^2-5x-3\\ 5x^2-x-11=0\)
a b c
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = {1 \pm \sqrt{1-4\cdot 5\cdot (-11)} \over 2\cdot 5}\)
\(x=\frac{1\pm\sqrt{221}}{10}=0.1\pm 1.48660687 \)
\(x_1=1,58660687\)
\(x_2=-1.38660687\)
!