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:

$a=2$

$b=-5$

$c=-3$

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

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

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

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

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

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

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

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