Question 1161300
Solution.
As the condition given
t represents time in years
P represents population
(a)Time in 2020 = 2020-2000 = 20
P = 6t^2+110t+4000
P = 6(20)^2+(110*20)+4000
Evaluate the power
P= 6*400+(110*20)+4000
Multiply the numbers
P = 2400+2200+4000
Add the numbers
P = 8600
Population will be 8600 in 2020.
(b)When the population be 6000
P = 6t^2+110t+4000
Use substitution
6000 = 6t^2+110t+4000
Move expression to the left side and change its sign
6000-6t^2-110t-4000 = 0
Subtract the terms
2000-6t^2-110t = 0
Use the commutative property to reorder the terms
-6t^2-110t+2000 = 0
Divide both sides of the equation by -2
3t^2+55t-1000 = 0
Solve the quadratic equation
ax^2+bx+c = 0 using
x = (-b±√(b²-4ac))/2a
t = (-55±√(55²-4*3*-1000))/2*3
t = (-55±√15025)/6
simplify the root
t = (-55±5√601)/6
Separate the solutions
t₁ = (-55+5√601)/6
t₂ = (-55-5√601)/6
Alternative form
t₁ = -29.59608
t₂ = 11.26275
Since negative value is not possible for the time in years
t = 11.26275