Answer:
[tex]x=\frac{1+ \sqrt{47}i}{8}[/tex] and [tex]x=\frac{1 -\sqrt{47}i}{8}[/tex]
Step-by-step explanation:
Given quadratic equation is [tex]4x^2-x+3=0[/tex].
Compare with [tex]ax^2+bx+c=0[/tex], we get:
a=4, b=-1, c=3
Now plug these values into quadratic formula :
[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-\left(-1\right) \pm \sqrt{\left(-1\right)^2-4\left(4\right)\left(3\right)}}{2\left(4\right)}[/tex]
[tex]x=\frac{1 \pm \sqrt{1-48}}{8}[/tex]
[tex]x=\frac{1 \pm \sqrt{-47}}{8}[/tex]
[tex]x=\frac{1 \pm \sqrt{47}i}{8}[/tex]
Hence final answer is
[tex]x=\frac{1+ \sqrt{47}i}{8}[/tex] and [tex]x=\frac{1 -\sqrt{47}i}{8}[/tex]
