How do you find the roots, real and imaginary, of $y = 2 x^{2} + 5 x - 12$ $y=2x^2 + 5x - 12$ using the quadratic formula?

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Answer 1 · 2016-02-25T06:10:07

Look at $(b^{2} - (4 \times a \times c))$ $(b^2-(4xx a xx c))$ part of the formula.

In a quadratic equation the formula to find the roots is

$x = \frac{- b \pm \sqrt{b^{2} - (4 \times a \times c)}}{2 a}$ $x=(-b+-sqrt(b^2-(4xx a xx c)))/(2a)$

If $- \sqrt{b^{2} - (4 \times a \times c)}$ $-sqrt(b^2-(4xx a xx c))$ is positive, the roots of the given function are real.

If $(b^{2} - (4 \times a \times c))$ $(b^2-(4xx a xx c))$ is negative, the roots of the given function are imaginary.

In our case -

$(5^{2} - (4 \times 2 \times (- 12)$ $(5^2-(4 xx 2 xx (-12)$
$(25 - (- 96)$ $(25-(-96)$
$(25 + 96) = 121 > 0$ $(25+96)=121>0$

Since $(b^{2} - (4 \times a \times c)) = 121 > 0$ $(b^2-(4xx a xx c))=121>0$ positive, the roots of the given function are real.

How do you find the roots, real and imaginary, of y=2x^2 + 5x - 12 y=2x2+5x−12y=2x^2 + 5x - 12 using the quadratic formula?