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Answer:
[tex]\textbf{A. }x=\dfrac{-4\pm2\sqrt{13}}{3}[/tex]
Step-by-step explanation:
First of all, subtract 12 from both sides of the equation to put it into standard form:
3x² +8x -12 = 0
Then identify the a, b, c of this form:
ax² +bx +c = 0
You see that a=3, b=8, c=-12.
Now use those values in the quadratic formula for the solutions:
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-8\pm\sqrt{8^2-4(3)(-12)}}{2(3)}\\\\x=\dfrac{-8\pm\sqrt{208}}{6}=\dfrac{-8\pm4\sqrt{13}}{6}\\\\\boxed{x=\dfrac{-4\pm2\sqrt{13}}{3}}\qquad\text{factor 2 from numerator and denominator}[/tex]
