Question

What values of x would make the following equation true?

Answer 1

There are many ways to do this. Such ways are using the quadratic formula, completing the square, etc.
We have $x^2+10x-21=0$ We can add 46 to both sides so the -21 becomes a +25. by doing that, we get $x^2+10x+25=46$ This can be wrote as $(x+5)^2=46$ We solve the square root and get $(x+5)= +-\sqrt{46}$ (The +- is a plus minus sign.) Subtract five on both sides to get $x=-5+- \sqrt{46}$
The answer is C.

Answer 2

Here we want to find the two solutions of a quadratic equation, to do it, we will use the Bhaskara's formula.

The correct option is C.

$x = -5 \pm \sqrt{46}$

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First, let's define the Bhaskara's formula.

For a general quadratic equation:

$a*x^2 + b*x + c = 0$

The solutions are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4*a*c} }{2*a}$

In this case, the quadratic equation is:

$x^2 + 10x - 21 = 0$

If we apply the Bhaskara's formula we get:

$x = \frac{-10 \pm \sqrt{(10)^2 - 4*1*(-21)} }{2*1} \\\\x = \frac{-10\pm \sqrt{184} }{2} \\\\x = \frac{-10}{2} \pm \frac{\sqrt{184} }{\sqrt{4} } \\\\x = -5 \pm \sqrt{46}$

Then we can see that the correct option is C.

If you want to learn more, you can read:

https://brainly.com/question/17177510