Question 1025573
The current through a particular wire can be modeled by
{{{I(t) = 2t^2-6t+7}}} 
where {{{I(t)}}} is the current in amperes after {{{t}}} seconds.
 
This is another one of those "plug-numbers-into-the-equation" problems.
 
a) Substituting {{{0.5}}} for {{{t}}} , we get
{{{I(0.5) = 2*0.5^2-6*0.5+7=2*0.25-3+7=0.5-3+7=4.5}}} 
The current in the wire after 0.5 seconds would be about {{{highlight(4.5)}}} amperes. 
 
(b) {{{I(t)=14}}} means {{{2t^2-6t+7=14}}}
{{{2t^2-6t+7=14}}}-->{{{2t^2-6t+7-14=0}}}-->{{{2t^2-6t-7=0}}} .
That quadratic equation can be solved by using the quadratic formula or by completing the square.
The quadratic formula says that the solutions to {{{ax^2+bx+c=0}}} (if any)
are given by {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} .
In our case, {{{system(x=t,a=2,b=6,c=-7)}}} , so the solution comes from
{{{t = (-(-6) +- sqrt((-6)^2-4*2*(-7)))/(2*2) = (6 +- sqrt(36+56))/4 = (6 +- sqrt(92))/4}}}={{{system(t=about-0.9,"or",t=about3.8979)}}}.
The negative value for {{{t}}} makes no sense, because the problem does not tell us what happens for {{{t<0}}}; it is not part of the domain of the function; probably because {{{t=0}}} was probably the moment the circuit was closed and current started flowing through the wire.
Rounding the positive answer to two decimal places, we get {{{t=3.90}}} .
In part (a), we easily included units with your numerical answers, because the units (amperes) were already written for us after the blank space.
Here it is not so, but we know that {{{t}}} is measured in seconds and {{{I(t)}}} is measured in amperes.
After about {{{highlight("3.90 seconds")}}} the current would reach 14 amperes. 
 
(c) Similarly to what was done for (b), {{{I(t)=60}}} means {{{2t^2-6t+7=60}}}
{{{2t^2-6t+7=60}}}-->{{{2t^2-6t+7-60=0}}}-->{{{2t^2-6t-53=0}}} .
Using the quadratic formula,
{{{t = (-(-6) +- sqrt((-6)^2-4*2*(-53)))/(2*2) = (6 +- sqrt(36+424))/4 = (6 +- sqrt(460))/4}}} .
The positive solution, rounded to  two decimal places is {{{t=6.86}}} .
After about {{{highlight("6.86 seconds")}}} the current would reach 60 amperes.