Answer:
c) x = 1 ± i √29 / 5
Step-by-step explanation:
assuming we are solving for x we can use the quadratic formula
quadratic formula : [tex]\frac{-b+or-\sqrt{b^2-4(a)(c)} }{2(a)}[/tex]
where the values of a,b and c are derived from the equation
the equation is put in quadratic form : ax² + bx + c
so in 5x² -2x + 6 a = 5 , b = - 2 and c = 6
we then plug these values of a, b and c into the quadratic formula
recall quadratic formula [tex]\frac{-b+or-\sqrt{b^2-4(a)(c)} }{2(a)}[/tex]
a = 5 , b = - 2 , c = 6
[tex]\frac{-(-2)+or-\sqrt{(-2)^2-4(5)(6)} }{2(5)}[/tex]
remove parenthesis at -(-2) to get positive 2 because the negative signs cancel out
[tex]\frac{2+or-\sqrt{(-2)^2-4(5)(6)} }{2(5)}[/tex]
evaluate exponent
[tex]\frac{2+or-\sqrt{4-4(5)(6)} }{2(5)}[/tex]
multiply -4(5)(6)
[tex]\frac{2+or-\sqrt{4-120} }{2(5)}[/tex]
subtract 120 from 4
[tex]\frac{2+or-\sqrt{-116} }{2(5)}[/tex]
multiply 2 and 5
[tex]\frac{2+or-\sqrt{-116} }{10}[/tex]
because we cant have a negative square root we replace the negative sign with i (which equals -1 ) and add it to the outside of the square root
[tex]\frac{2+or-i\sqrt{116} }{10}[/tex]
we then simplify the radical
[tex]\frac{2+or-2i\sqrt{29} }{10}[/tex]
we then simplify the fraction
2/10 = 1/5 so both 2's cancel out
[tex]\frac{1+or-i\sqrt{29} }{10}[/tex]
the answer is C
