Fixed typos (#27202)

This commit is contained in:
Andrew Mackie
2019-03-07 18:46:51 -08:00
committed by Randell Dawson
parent edb3548826
commit d37f5a7751

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@ -21,14 +21,14 @@ A quadratic equation generally takes the form: <em>ax<sup>2</sup> + bx + c</em>
4. Completing the square on the Left Hand Side and simplifying the Right Hand Side of the above equation, we have: 4. Completing the square on the Left Hand Side and simplifying the Right Hand Side of the above equation, we have:
<pre>(x<sup></sup> + <sup>b</sup>/<sub>2a</sub>)<sup>2</sup> = (<sup>b<sup>2</sup></sup>/<sub>4a<sup>2</sup></sub>) - (<sup>c</sup>/<sub>a</sub>)</pre> <pre>(x<sup></sup> + <sup>b</sup>/<sub>2a</sub>)<sup>2</sup> = (<sup>b<sup>2</sup></sup>/<sub>4a<sup>2</sup></sub>) - (<sup>c</sup>/<sub>a</sub>)</pre>
5. Further simplpfying the Right Hand Side, 5. Further simplifying the Right Hand Side,
<pre>(x<sup></sup> + <sup>b</sup>/<sub>2a</sub>)<sup>2</sup> = (b<sup>2</sup> - 4ac)/4a<sup>2</sup> </pre> <pre>(x<sup></sup> + <sup>b</sup>/<sub>2a</sub>)<sup>2</sup> = (b<sup>2</sup> - 4ac) &divide; 4a<sup>2</sup> </pre>
6. Finding the square root of both sides of the equation, 6. Finding the square root of both sides of the equation,
<pre>x<sup></sup> + <sup>b</sup>/<sub>2a</sub> = &radic;(b<sup>2</sup> - 4ac) &divide; 2a </pre> <pre>x<sup></sup> + <sup>b</sup>/<sub>2a</sub> = &pm;((b<sup>2</sup> - 4ac)<sup>&half;</sup> &divide; 2a) </pre>
7. By making x the subject of our formula, we are able to solve for its value completely: 7. By making x the subject of our formula, we are able to solve for its value completely:
<pre>x<sup></sup> = -b &#177; &radic;(b<sup>2</sup> - 4ac) &divide; 2a </pre> <pre>x<sup></sup> = (-b &pm; (b<sup>2</sup> - 4ac)<sup>&half;</sup>) &divide; 2a </pre>
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