Question 177030
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Edwin's solution follows.  You do not need the quadratic formula, as the resulting quadratic factors:
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{{{matrix(1,3,abs(matrix(3,3,4,x,-2,-x,-3,1,-6,2,3)), "=", -3)  ) }}}
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First you have to learn how to find the value of
a 3x3 determinant:

Suppose you have the 3x3 determinant:

{{{abs(matrix(3,3,A,B,C,D,E,F,G,H,I))}}}

Copy the first two columns over on the right:

{{{abs(matrix(3,3,A,B,C,D,E,F,G,H,I))matrix(3,2,A,B,D,E,G,H)}}}

Then add the products of the three diagonals that go down to the
RIGHT:  +AEI+BFG+CDH

Then subtract the products of the three diagonals that go down to the
LEFT:  -CEG-AFH-BDI

So 

{{{matrix(1,3,abs(matrix(3,3,A,B,C,D,E,F,G,H,I))matrix(3,2,A,B,D,E,G,H), "=", AEI+BFG+CDH-CEG-AFH-BDI)) }}}


{{{matrix(1,3,abs(matrix(3,3,4,x,-2,-x,-3,1,-6,2,3)), "=", -3)  ) }}}

Evaluate the determinant according to the rules 
above or whatever rules you learned. There are more
than one possible set of rules for this.  Use 
whatever method you have studied.  If you use the 
method above, you'll copy the first two columns over:

{{{matrix(1,3,abs(matrix(3,3,4,x,-2,-x,-3,1,-6,2,3))matrix(3,2,4,x,-x,-3,-6,2), "=", -3)  ) }}}

Then

{{{matrix(9,3,

(4)(-3)(3)+(x)(1)(-6)+(-2)(-x)(2)-(-2)(-3)(-6)-(4)(1)(2)-(x)(-x)(3), "=",-3,
-36-6x+4x+36-8+3x^2,"=", -3,
-2x-8+3x^2,"=",-3,
3x^2-2x-8,"=",-3,
3x^2-2x-5,"=",0,
(3x-5)(x+1),"=",0,
3x-5=0,x+1=0,"",
3x=5,x=-1,"",
x=5/3,"","") }}}

So there are two solutions, {{{x=5/3}}} and {{{x=-1}}}

Edwin</pre>