How do you find the roots, real and imaginary, of `y= 5x^2 - 13x -4-x(x-1)` using the quadratic formula?

Here,

`y=f(x)=5x^2-13x-4-x(x-1)`

`f(x)=5x^2-13x-4-x^2+x`

`f(x)=4x^2-12x-4`

We have to represents `(4x^2-12x-4)` as a quadratic
equation,before using quadratic formula.

Let, `4x^2-12x-4=0`

Comparing with `ax^2+bx+c=0,` we get

`a=4,b=-12 and c=-4`

Now, `triangle=b^2-4ac=144-4(4)(-4)=208 > 0`

Hence,

`triangle > 0=>"Both the roots are "color(red)"REAL and Different"`

Note:

`(1)triangle > 0=>"Both the roots are "color(red)"REAL and Different"`

`(2)triangle = 0=>"Both the roots are "color(red)"REAL and Equal"`

`(3)triangle < 0=>"Both the roots are "color(red)"Imaginary and Different"` `color(white)(,................................................)`.`""color(blue)"conjugate complex root"`

`y = 5 x^2 -13 x - 4 - x(x-1)` or

`y = 5 x^2 -13 x - 4 - x^2 +x` or

`y = 4 x^2 -12 x - 4` or

`y = 4( x^2 -3 x - 1)` Comparing with standard quadratic

equation `ax^2+bx+c=0; a=1 ,b=-3 ,c=-1`

Discriminant `D= b^2-4 a c = 9+4=13`

Discriminant positive, we get two real roots,

Quadratic formula: `x= (-b+-sqrtD)/(2a)`or

`x= (3+-sqrt 13)/2 = 3/2 +- sqrt 13/2`

`y = 4(x- ( 3/2 + sqrt 13/2))(x- ( 3/2 - sqrt 13/2))`

Zeros are `x = 3/2 +- sqrt 13/2`

[Ans]