Posted by gabriel on Tuesday, June 8, 2010 at 8:42am.
I would use the standard quadratic formula for ax²+bx+c=0, and
x=(-b±√(b²-4ac))/2a
Here
4x^2 - 5x = -2
can be rewritten as
4x^2 - 5x + 2 =0, so
a=4,
b=-5
c=2
substituting in the quadratic formula,
x=(-(-5)±√((-5)²-4(4)(2)))/(2*4)
which when simplified, gives
x=(5±(√7)i)/8 where i=√(-1) (complex root).
Complex numbers is a little beyond high school, but the procedure is the same.
If you are not expecting complex roots, you may want to recheck if the question you posted was correct. If not, repeat the exercise using the correct question.
using the quadratic formula, how do you solve 2p^2 + 16p = -2p^2 + 3?
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