Question 134509
Given: {{{T=5.8t^2+13t+32}}}
a) What is the car's radiator's temperature at the instant the car is turned on? What is the time when the car is turned on? time = 0
So replace t with 0 in the function and solve
{{{T=5.8*0^2+13*0+32}}} 
{{{T = 32}}}

b) What is the car's radiator's temperature after the car has been driven for 3 minutes? 
What value should yo use for t now? t = 3
{{{T=5.8*3^2+13*3+32}}} 
{{{T = 5.8*9 + 13*3 + 32}}}
{{{T = 123.2}}}

c) How long after the car has begun operating will the car's radiator temperature reach 90 degrees Fahrenheit?  
This time you are given a tempurature (T) and asked to find the time.
{{{90=5.8t^2+13t+32}}}
{{{0 = 5.8t^2 + 13t - 58}}}

Use the quadratic equation to find values for t. Then pick the value that 'makes sense"
0 = at^2 + bt + c, so a = 5.8, b = 13 and c = -58
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

I get t = 2.24

Check your answer by replacing t with 2.23
{{{T = 5.8* 2.24^2 + 13 * 2.24 + 32}}}
{{{T = 29.01 + 29.12 + 32 }}}  ~ 90