How do you use the quadratic formula to solve $3 x^{2} - 2 x + 7 = 0$ $3x^2-2x+7=0$ ?

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How do you use the quadratic formula to solve $3 x^{2} - 2 x + 7 = 0$ $3x^2-2x+7=0$ ?

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Answer 1 · 2015-02-04T22:03:11

Our equation is of the form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , therefore, we can use the quadratic formula. The quadratic formula is $x = \frac{- b \pm \sqrt{D}}{2 a}$ $x = (-b +- sqrt(D))/(2a)$ , with $D = b^{2} - 4 a c$ $D = b^2-4ac$ . We see that $a = 3, b = - 2$ $a=3, b=-2$ and $c = 7$ $c=7$ .

Firstly, we calculate $D$ $D$ .
$D = b^{2} - 4 a c = {(- 2)}^{2} - 4 \cdot 3 \cdot 7 = - 80$ $D = b^2-4ac = (-2)^2-4*3*7 = -80$ . Because D is less than zero, there are no real roots. We can however calculate the imaginary solutions. We just calculate $x$ $x$ :
$x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{2 \pm \sqrt{- 80}}{2 \cdot 3} = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{16} \cdot \sqrt{5}}{6} = \frac{2 \pm i \cdot 4 \sqrt{5}}{6} = \frac{1}{3} \pm \frac{2}{3} i \sqrt{5}$ $x = (-b +- sqrt(D))/(2a) = (2+-sqrt(-80))/(2*3) = (2+-sqrt(-1)*sqrt(16)*sqrt(5))/6 = (2+-i*4sqrt(5))/6 = 1/3+- 2/3 isqrt(5)$

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How do you use the quadratic formula to solve $3 x^{2} - 2 x + 7 = 0$ $3x^2-2x+7=0$ ?

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