Solve for x. (2x - 1)(3x + 2) = x + 292
Expand the LHS and then move the RHS to the LHS and simplify. There is still more to do.
Expand means to multiply out.
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are you given options?
no
The following is from the Mathematica program and the terms are not in classroom order. \[6 x^2+x-(x+292)-2=0 \]\[6 x^2-294=0\]The above can be simplified and factored.
than you have to multiply the LHS and solve for the value of x.
(2x-1)(3x-2) =x+292 6x^2 -4x-3x +2 =x+292 6x^2 -8x -290 =0 3x^2 -4x -145 =0 so just solve this quadratic for x
@jhonyy9 You did not copy the problem as given, you changed the sign.
\[6 x^2-294=0 \]\[x^2-49=0 \]
yes right there is a sign inversed what i changed - sorry
yes, robtobey has resolved it down to the difference of two perfect squares, where it is easy to solve by factoring.
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