To solve a quadratic equation by factoring:
EXAMPLES:
Solve:
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$3x^2 = 5  14x$
Solution:
Write a nice, clean list of equivalent equations:
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$3x^2 = 5  14x$

(original equation) 
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$3x^2 + 14x  5 = 0$ 
(put in standard form: subtract $\,5\,$ from both sides; add $\,14x\,$ to both sides) 
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$(3x1)(x+5) = 0$ 
(factor the lefthand side; you may want to use the factor by grouping method) 
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$3x1 = 0\ \ \text{ or }\ \ x + 5 = 0$ 
(use the Zero Factor Law) 
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$3x = 1\ \ \text{ or }\ \ x = 5$ 
(solve the simpler equations) 
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$\displaystyle x = \frac{1}{3}\ \ \text{ or }\ \ x = 5$ 
(solve the simpler equations) 
Check by substituting into the original equation:
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$3{(\frac{1}{3})}^2 \ \ \overset{\text{?}}{=}\ \ 5  14(\frac13)\,$; $\,3\cdot\frac19 \overset{\text{?}}{=} \frac{15}{3}\frac{14}3\,$; $\,\frac13 = \frac 13$; Check!
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$3{(5)}^2 \ \ \overset{\text{?}}{=}\ \ 5  14(5)\,$; $\,3\cdot 25 \ \overset{\text{?}}{=}\ 5 + 70\,$; $\,75 = 75$; Check!
Solve:
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$(2x+3)(5x1) = 0$
Solution:
Note: Don't multiply it out!
If it's already in factored form, with zero on one side,
then be happy that a lot of the work has already been done for you.
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$(2x+3)(5x1) = 0$

(original equation) 
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$2x+3 = 0\ \ \text{ or }\ \ 5x  1 = 0$ 
(use the Zero Factor Law) 
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$2x = 3\ \ \text{ or }\ \ 5x = 1$ 
(solve the simpler equations) 
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$\displaystyle x = \frac{3}{2}\ \ \text{ or }\ \ x = \frac{1}{5}$ 
(solve the simpler equations) 
Check by substituting into the original equation:
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$(2(\frac32)+3)(5(\frac32)1) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!
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$(2(\frac15)+3)(5(\frac15)1) \ \ \overset{\text{?}}{=}\ \ 0\,$; $\,0 = 0\,$; Check!
Solve:
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$10x^2  11x  6 = 0$
Solution:
Note that it's already in standard form.
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$10x^2  11x  6 = 0$

(original equation) 
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$(5x+2)(2x3) = 0$ 
(factor the lefthand side; you may want to use the factor by grouping method) 
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$5x+2 = 0\ \ \text{ or }\ \ 2x  3 = 0$ 
(use the Zero Factor Law) 
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$5x = 2\ \ \text{ or }\ \ 2x = 3$ 
(solve the simpler equations) 
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$\displaystyle x = \frac{2}{5}\ \ \text{ or }\ \ x = \frac{3}{2}$ 
(solve the simpler equations) 
Check by substituting into the original equation:
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$10(\frac25)^2  11(\frac25)  6 \ \ \overset{\text{?}}{=}\ \ 0\,$;
$10(\frac{4}{25}) + \frac{22}{5}  6 \ \ \overset{\text{?}}{=}\ \ 0\,$;
$2(\frac{4}{5}) + \frac{22}{5}  \frac{30}{5} \ \ \overset{\text{?}}{=}\ \ 0\,$;
$\,0 = 0\,$; Check!
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$10{(\frac32)}^2  11(\frac32)  6 \ \ \overset{\text{?}}{=}\ \ 0\,$;
$10(\frac{9}{4})  \frac{33}{2}  6 \ \ \overset{\text{?}}{=}\ \ 0\,$;
$5(\frac{9}{2})  \frac{33}{2}  \frac{12}{2} \ \ \overset{\text{?}}{=}\ \ 0\,$;
$\,0 = 0\,$; Check!