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The book says there should be a quadratic formula, but I don't understand how to solve it.

I have: $\pm 6.0 = \sqrt(\frac{36 - 4(4.0)( \pm 7.04)}{14.2})$

I don't understand how to input this into a calculator or solve this to get the final answer.

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  • $\begingroup$ It's difficult to decipher what expression you have written there. $\endgroup$ – ervx Feb 15 '17 at 3:53
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The symbol $\pm$ ("$+$ or $-$") means that either $+$ or $-$ will produce a solution - so use each, getting a different number each time, and you will get two answers.

For example: say I've solved an equation and gotten $x = 2 \pm 7$. Then $2 - 7 = -5$ and $2 + 7 = 9$ are both solutions to my equation. What that means is that if I've only been asked for "a" solution, I can give either $-5$ or $9$ as my answer - both are correct. But if I've been asked for "the" solution or "all" solutions, then the correct answer is "$x = -5$ or $x = 9$".

Your particular example looks like $\frac{6 \pm \sqrt{36 - 4 \cdot 4.0 \cdot \pm 7.04}}{14.2}$. If that's correct, then there are two $\pm$'s to deal with, and these can be evaluated independently. For example, if I were looking at $1 \pm 2 \pm 3$, then there would be four possible solutions: $1 + 2 + 3$, $1 + 2 - 3$, $1 - 2 + 3$, and $1 - 2 - 3$.

Word of warning: I said "if that's correct" because it doesn't look like it is. You referenced the quadratic formula, but if that's the only place $\pm$'s are coming from, there shouldn't be two of them. The quadratic formula states that the solutions to the quadratic $ax^2 + bx + c = 0$ are $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$; notice that there's only one $\pm$ in that.

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I can't really make out what the equation is but basically the quadratic equation gives you the solutions of $ax^{2}+bx+c=0$:

$$ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. $$

There is a $\pm$ because you may have two solutions; namely,

$$ x_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\qquad\text{ and }\qquad x_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}. $$

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