Answer: [tex]-7.60,\ -4.47[/tex]
Step-by-step explanation:
Given
[tex]f(x)=-1.3x^2-15.7x-44.2[/tex]
The solution of the standard quadratic equation [tex]ax^2+bx+c=0[/tex] has solution
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Applying the above principle
[tex]\Rightarrow x=\dfrac{15.7\pm\sqrt{(-15.7)^2-4(-1.3)(-44.2)}}{2(-1.3)}\\\\\Rightarrow x=\dfrac{15.7\pm\sqrt{246.49-229.84}}{-2.6}\\\\\Rightarrow x=\dfrac{15.7\pm \sqrt{16.65}}{-2.6}\\\\\Rightarrow x=\dfrac{15.7\pm 4.08}{-2.6}\\\\\Rightarrow x=-7.60,\ -4.47[/tex]
The zeroes of the equation are -7.60, -4.47.
