Question 729504
you did good job, now just use quadratic formula to solve for {{{x}}}

{{{5x^2+3x+1=0}}}..note that coefficient {{{a=5}}}, {{{b=3}}} and {{{c=1}}} 

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...plug coefficients in

{{{x = (-3 +- sqrt( 3^2-4*5*1 ))/(2*5) }}}


{{{x = (-3 +- sqrt( 9-20))/10 }}}

{{{x = (-3 +- sqrt( -11))/10 }}}......you have sqrt of negative number and that means there is no real solutions only imaginary and graph will not intersect with x-axis, and solutions are complex numbers  

{{{x = (-3 +- sqrt( -11))/10 }}}......write {{{sqrt(-11)}}} as {{{sqrt(-1*11)}}}


 {{{x = (-3 +- sqrt( -1*11))/10 }}}...since {{{sqrt(-1)=i}}} we have


 {{{x = (-3 +- sqrt(11)i)/10 }}}


{{{x = (-3 +- 3.32i)/10 }}}

solutions:

{{{x = (-3 + 3.32i)/10 }}}

{{{x = -3/10 + 3.32i/10 }}}

or


{{{x = (-3 - 3.32i)/10 }}}

{{{x = -3/10 -3.32i/10 }}}


{{{ graph( 600, 600, -5,5, -5, 10,5x^2+3x+1) }}}