How do you find the roots, real and imaginary, of $y = 3 x^{2} - 2 x + 12$ $y=3x^2-2x+12$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = 3 x^{2} - 2 x + 12$ $y=3x^2-2x+12$ using the quadratic formula?

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Answer 1 · 2016-01-01T15:00:56

First, equate to $0$ $0$

ie. $3 x^{2} - 2 x + 12 = 0$ $3x^2 - 2x +12 = 0$

This is of the form $a x^{2} + b x + c$ $ax^2 + bx + c$ , where $a = 3$ $a=3$ , $b = - 2$ $b=-2$ and $c = 12$ $c=12$ .

Now use the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b±sqrt(b^2-4ac))/(2a)$

Substitute values into formula:

$x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (3) (12)}}{2 (3)} = \frac{2 \pm \sqrt{4 - 144}}{6} = \frac{2 \pm \sqrt{- 100}}{6} = \frac{2 \pm 10 i}{6} = \frac{1 \pm 5 i}{3} = \frac{1}{3} \pm \frac{5}{3} i$ $x=(-(-2)±sqrt((-2)^2-4(3)(12)))/(2(3))=(2±sqrt(4-144))/6=(2±sqrt(-100))/6=(2±10i)/6=(1pm5i)/3=1/3pm5/3i$

Roots are $\frac{1}{3} + \frac{5}{3} i$ $1/3+5/3i$ and $\frac{1}{3} - \frac{5}{3} i$ $1/3-5/3i$ .

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How do you find the roots, real and imaginary, of $y = 3 x^{2} - 2 x + 12$ $y=3x^2-2x+12$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = 3 x^{2} - 2 x + 12$ $y=3x^2-2x+12$ using the quadratic formula?