Question

Find the zeros of the function f(x)=-1.3x^2-15.7x-44.2 . Round values to the nearest hundredth (if necessary).

Answer 1

Answer:
`-7.60,\ -4.47`

Explanation:

Given

`f(x)=-1.3x^2-15.7x-44.2`

The solution of the standard quadratic equation
`ax^2+bx+c=0` has solution

`x=(-b\pm√(b^2-4ac))/(2a)`

Applying the above principle

`\Rightarrow x=(15.7\pm√((-15.7)^2-4(-1.3)(-44.2)))/(2(-1.3))\\\\\Rightarrow x=(15.7\pm√(246.49-229.84))/(-2.6)\\\\\Rightarrow x=(15.7\pm √(16.65))/(-2.6)\\\\\Rightarrow x=(15.7\pm 4.08)/(-2.6)\\\\\Rightarrow x=-7.60,\ -4.47`

The zeroes of the equation are -7.60, -4.47.