Proximate the real zeros of f(x)=x^2+5x+3 to the nearest tenth The zeros are about and about

x = sqrt(13)/2 - 5/2 or x = -5/2 - sqrt(13)/2

or decimal -0.69722436226800535344038936626475202687435171307737689364

Step-by-step explanation:

Solve for x:

x^2 + 5 x + 3 = 0

Subtract 3 from both sides:

x^2 + 5 x = -3

Add 25/4 to both sides:

x^2 + 5 x + 25/4 = 13/4

Write the left hand side as a square:

(x + 5/2)^2 = 13/4

Take the square root of both sides:

x + 5/2 = sqrt(13)/2 or x + 5/2 = -sqrt(13)/2

Subtract 5/2 from both sides:

x = sqrt(13)/2 - 5/2 or x + 5/2 = -sqrt(13)/2

Subtract 5/2 from both sides:

 x = sqrt(13)/2 - 5/2 or x = -5/2 - sqrt(13)/2

an = 22 + (n-1) *6

110 seats

step-by-step explanation:

an=a1+(n-1)(d)

a1 = 22   this is the first item in the sequence

d =6   this is how the sequence goes up

a)   an = 22 + (n-1) *6

b) a12 = 22 + (12-1) *6

    a12 = 22+11*6

  a12 = 22 +88

  a12 = 110