What are the solutions of #x^2-3x=-10#?

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Answer 1 · 2015-10-22T18:11:55

The solutions are # 3/2 pm i *sqrt(31)/2#, where #i=sqrt{-1}# is the imaginary unit.

Write the equation in the form #a x^2 +bx + c=0#: #x^2-3x=-10 implies x^2-3x+10=0#.

The solutions, by the quadratic formula, are then:

#x=(-b pm sqrt(b^2-4ac))/(2a)=(3 pm sqrt(9-4*1*10))/(2*1)#

#=(3 pm sqrt(-31))/2 = 3/2 pm i *sqrt(31)/2#, where #i=sqrt{-1}# is the imaginary unit.

Answer 2 · 2015-10-22T18:15:37

Imaginary numbers

=#x^2-3x+10=0#
graph{x^2-3x+10 [-23.11, 34.1, -3.08, 25.54]}
Using --- (b+ or - #sqrt(b^2-4ac)# )/ 2a

You'll see that it has complex roots, so by plotting a graph you can figure out the answer, but in the form

=
1.5 + 2.78388i
1.5 - 2.78388i