Answer:
[tex]\displaystyle x=\frac{-3+\sqrt{6}}{3}\text{ and } x=\frac{-3-\sqrt{6}}{3}[/tex]
Or, by approximating:
[tex]x\approx-0.1835\text{ or } x\approx -1.8165[/tex]
Step-by-step explanation:
Let’s convert this to standard form. We have:
[tex]-6x-1+5x^2=8x^2[/tex]
Subtract 5x² from both sides:
[tex]-6x-1=3x^2[/tex]
And add 6x and 1 to both sides:
[tex]0=3x^2+6x+1[/tex]
This is not factorable. So, we will need to use the quadratic formula.
The quadratic formula for a quadratic in standard form is given by:
[tex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}[/tex]
In this case, a=3; b=6, and c=1.
Substitute appropriately:
[tex]\displaystyle x=\frac{-6\pm\sqrt{6^2-4(3)(1)}}{2(3)}[/tex]
Simplify:
[tex]\displaystyle x=\frac{-6\pm\sqrt{24}}{6}[/tex]
We can simplify the square root:
[tex]\sqrt{24}=\sqrt{4}\cdot\sqrt{6}=2\sqrt{6}[/tex]
Hence:
[tex]\displaystyle x=\frac{-6\pm2\sqrt{4}}{6}[/tex]
Simplify:
[tex]\displaystyle x=\frac{-3\pm\sqrt{6}}{3}[/tex]
Hence, we will have two solutions:
[tex]\displaystyle x=\frac{-3+\sqrt{6}}{3}\text{ and } x=\frac{-3-\sqrt{6}}{3}[/tex]
Approximating them, we can see that our solutions are approximately:
[tex]x\approx-0.1835\text{ or } x\approx -1.8165[/tex]