That's a question about quadratic function.
Any quadratic function can be represented by the following form:
[tex]\boxed{f(x)=ax^2+bx+c}[/tex]
Example:
[tex]f(x)= -3x^2-9x+57[/tex] is a function where [tex]a=-3[/tex], [tex]b=-9[/tex] and [tex]c=57[/tex].
Okay, in our problem, we need to find the value of x when [tex]f(x)=0[/tex]. That's mean that the result of our function is equal to zero. Therefore, we have the quadratic equation below:
[tex]x^2+4x-5=0[/tex]
To solve a quadratic equation, we use the Bhaskara's formula. Do you remember the value of a, b and c? They going to be important right now. This is the Bhaskara's formula:
[tex]\boxed{x=\frac{-b\pm \sqrt{b^2-4ac} }{2a} }[/tex]
So, let's see the values of a, b and c in our equation and apply them in the Bhaskara's formula:
In [tex]x^2+4x-5=0[/tex] equation, [tex]a=1[/tex], [tex]b=4[/tex] and [tex]c=-5[/tex]. Let's replace those values:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
[tex]x=\frac{-4\pm \sqrt{4^2-4\times1\times(-5)} }{2\cdot1}[/tex]
[tex]x=\frac{-4\pm \sqrt{16-(-20)} }{2}[/tex]
[tex]x=\frac{-4\pm \sqrt{16 + 20)} }{2}[/tex]
[tex]x=\frac{-4\pm \sqrt{36} }{2}[/tex]
[tex]x=\frac{-4\pm 6 }{2}[/tex]
From now, we have two possibilities:
To add:
[tex]x_1 = \frac{-4+6}{2} \\x_1=\frac{2}{2} \\x_1=1[/tex]
To subtract:
[tex]x_2=\frac{-4-6}{2} \\x_2=\frac{-10}{2} \\x_2=-5[/tex]
Therefore, the result of our problem is: [tex]x_1 = 1[/tex] and [tex]x_2=-5[/tex].
I hope I've helped. ^^
Enjoy your studies. \o/
