How do you find the zeros, real and imaginary, of #y=2x^2-5x-3 #using the quadratic formula?

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Answer 1 · 2017-11-30T00:17:33

There are only #2# real solutions:

#x=6\quad,\quad x=-1#

First, we can calculate the discriminant to see how many real/imaginary answers there will be:

#b^2-4ac#

#\rightarrow (-5)^2-4(1)(-3)#

#\rightarrow 25+24#

#\rightarrow 49#

#\therefore# the equation will have #2# real answers. There are no imaginary solutions.

Now we can solve for those answers.

First set the RHS equal to #0#:

#2x^2-5x-3=0#

Now we can just plug the following values into the quadratic formula:

#a=2#

#b=-5#

#c=-3#

#x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}#

#\rightarrow x=\frac{5\pm\sqrt{(-5)^2-4(2)(-3)}}{2(2)}#

#\rightarrow x=\frac{5\pm\sqrt{25+24}}{2}#

#\rightarrow x=\frac{5\pm\sqrt{49}}{2}#

#\rightarrow x=\frac{5\pm7}{2}#

#\rightarrow x=\frac{5+7}{2}\quad,\quad x=\frac{5-7}{2}3#

#\rightarrow x=6\quad, \quad x=-1#