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Answer 1 · 2017-12-13T14:48:30

The discriminant is $- 28$ $-28$ , meaning there are no real solutions to the above function.

Given an equation in the form $a x^{2} + b x + c$ $ax^2+bx+c$ where $a \neq q 0$ $a\neq 0$ , the discriminant is given by $b^{2} - 4 a c$ $b^2-4ac$ .

If . . .

$b^{2} - 4 a c > 0,$ $b^2-4ac > 0,$ there are $2$ $2$ real solutions.

$b^{2} - 4 a c = 0,$ $b^2-4ac =0,$ there is $1$ $1$ real solution.

$b^{2} - 4 a c < 0$ $b^2-4ac < 0$ , there are no real solutions.

In this case, the discriminant is $10^{2} - 4 (- 4) (- 8)$ $10^2-4(-4)(-8)$ ,

$\Rightarrow 100 - 128 \Rightarrow - 28$ $\implies 100-128\implies -28$

$\therefore$ there are no real solutions.