Respuesta :

First, we have this...
[tex]9x+2=8x^2+6[/tex]
Subtract [tex]8x^2+6x[/tex] from both sides.
[tex]9x+2-\left(8x^2+6x\right)=8x^2+6x-\left(8x^2+6x\right)[/tex]
Simplify.
[tex]-8x^2+3x+2=0[/tex]
Formula:
[tex]x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\left(-8\right)2}}{2\left(-8\right)}[/tex]
Plug in the numbers..
[tex]\frac{-3+\sqrt{3^2-4\left(-8\right)\cdot \:2}}{2\left(-8\right)}[/tex]
Simplify...
[tex]=-\frac{\sqrt{73}-3}{16}[/tex]
We are going to do the same, just with a different operation...
[tex]\frac{-3-\sqrt{3^2-4\left(-8\right)\cdot \:2}}{2\left(-8\right)}[/tex]
Simplify.
[tex]=\frac{\sqrt{73}+3}{16}[/tex]

So, the solutions are...
[tex]x=-\frac{-3+\sqrt{73}}{16},\:x=\frac{\sqrt{73}+3}{16}[/tex]

Have a nice day! :)

Answer:

Sample Response/Explanation: No. The correct values of a, b, and c were substituted in, but the formula was simplified wrong. The 64 should be added so the radicand is 73. There should be 2 real roots.