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Sagot :
Answer:
2
Step-by-step explanation:
[tex]5x^2 +3x +2 = 0\\\\\text{Apply the quadratic formula,}~ x=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\\text{In this case,}~a= 5, ~ b = 3~ \text{and}~ c = 2\\\\ \\x=\dfrac{-3\pm \sqrt{3^2 -4\cdot 5 \cdot 2}}{2\cdot 5}\\\\~~~=\dfrac{-3 \pm \sqrt{9-40}}{10}\\\\~~~=\dfrac{-3 \pm \sqrt{-31}}{10}\\\\~~~=\dfrac{-3 \pm i\sqrt{31}}{10}\\\\\text{So, there are 2 solutions,}~ x=\dfrac{-3 + i\sqrt{31}}{10}~ \text{and} ~x=\dfrac{-3 -i\sqrt{31}}{10}[/tex]
Another way to find the number of solutions is by looking at the highest exponent(Degree), which indicates the number of solutions.
[tex]\\ \rm\rightarrowtail 5x^2+3x+2=0[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{-3\pm \sqrt{9-40}}{10}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{-3\pm\sqrt{-31}}{10}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{-3\pm\sqrt{31}i}{10}[/tex]
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