Answer:
The equation [tex]-4x^2+10x-8[/tex] does not have any real zeroes.
Step-by-step explanation:
The given equation is [tex]-4x^2+10x-8[/tex].
Let us compare it with standard quadratic equation [tex]ax^2+bx+c[/tex]
a = -4
b = 10
c = -8
The nature of zeroes is determined by Discriminant 'D'.
1. If D = 0, the quadratic equation has two equal real zeroes.
2. If D > 0, the quadratic equation has two unequal real zeroes.
3. If D < 0, the quadratic equation has two non-real or imaginary zeroes.
Formula for D is:
[tex]D=b^{2} -4ac[/tex]
Putting the values of a, b and c to find D:
[tex]\Rightarrow 10^2 - 4(-4)(-8)\\\Rightarrow 10^2 - 4(4)(8)\\\Rightarrow 100 - 4(32)\\\Rightarrow 100 - 128\\\Rightarrow D = -28[/tex]
Here, D is negative so the zeroes of this quadratic equation are imaginary.
Hence, no real zeroes for the given equation [tex]-4x^2+10x-8[/tex].
