Answer:
[tex]x=\frac{5(+/-)\sqrt{41}} {4}[/tex]
Step-by-step explanation:
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2x^{2} -5x+1=3[/tex]
equate to zero
[tex]2x^{2} -5x+1-3=0[/tex]
[tex]2x^{2} -5x-2=0[/tex]
so
[tex]a=2\\b=-5\\c=-2[/tex]
substitute in the formula
[tex]x=\frac{-(-5)(+/-)\sqrt{-5^{2}-4(2)(-2)}} {2(2)}[/tex]
[tex]x=\frac{5(+/-)\sqrt{41}} {4}[/tex]
