ladiesman217
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ladiesman217
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\[x ^{2}4x=5\]

ajprincess
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\(x^24x=5\)
\(x^24x5=\)
use quadratic formula to find x.

Compassionate
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Korra stahp. Itachi is at work.

ajprincess
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who is itachi @compassionate?

Compassionate
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\[ \begin{array}l\color{red}{\text{D}}\color{orange}{\text{O}}\color{#E6E600}{\text{ }}\color{green}{\text{Y}}\color{blue}{\text{O}}\color{purple}{\text{U}}\color{purple}{\text{ }}\color{red}{\text{E}}\color{orange}{\text{V}}\color{#E6E600}{\text{E}}\color{green}{\text{N}}\color{blue}{\text{ }}\color{purple}{\text{A}}\color{purple}{\text{N}}\color{red}{\text{I}}\color{orange}{\text{M}}\color{#E6E600}{\text{E}}\color{green}{\text{?}}\color{blue}{\text{}}\end{array} \]

austinL
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\[x=\frac{b \pm \sqrt{b^24 a c}}{2a}\]

ajprincess
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the values of a, b nd c can be found by comparing the equation \(x^24x5=0\) with the general equation \(ax^2+bx+c=0\). Does that help? @ladiesman217

ladiesman217
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ok thnx

ladiesman217
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so x=1 & x=5

ajprincess
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plz check the signs:)

ladiesman217
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aren't those the correct signs

ajprincess
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nope:(

ajprincess
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what are the values of a, b and c?

ladiesman217
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\[1^{2}4*(1)=5\]

ajprincess
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oops reallly soorrry.u r 100% right:)