Answer:
(D) [tex]\begin{array}{*{20}c} {x=\frac{{ 5 \pm \sqrt {13} }}{{2}}} \end{array}[/tex]
Step-by-step explanation:
The quadratic formula is:
[tex]\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}[/tex]
Assuming that a is our x² term, b is our x term, and c is the constant, we can substitute inside the equation.
[tex]\begin{array}{*{20}c} {\frac{{ - (-5) \pm \sqrt {5^2 - 4\cdot1\cdot3} }}{{2\cdot1}}} \end{array}[/tex]
[tex]\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 12} }}{{2}}} \end{array}[/tex]
[tex]\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {13} }}{{2}}} \end{array}[/tex]
So the answer is D, [tex]\begin{array}{*{20}c} {x=\frac{{ 5 \pm \sqrt {13} }}{{2}}} \end{array}[/tex].
Hope this helped!
