How do you solve $0 = 3 x^{2} - 2 x - 3$ $0=3x^2-2x-3$ using the quadratic formula?

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Answer 1 · 2017-06-28T11:32:10

$X = 1.387, - 0.720$ $X = 1.387 , -0.720$

$D = b^{2} - 4 a c$ $D = b^2 - 4ac$

$\to {(- 2)}^{2} - 4 \times 3 \times - 3$ $-> (-2)^2 -4 xx 3 xx -3$

$D = 40$ $D = 40$

Now..

$X = \frac{(- b) \pm \sqrt{D}}{2 a}$ $X = ((-b) +- sqrt D)/(2a)$

$\to X = \frac{- (- 2) \pm \sqrt{40}}{2 \times 3}$ $-> X = (-(-2) +- sqrt 40)/(2×3)$

$:. x = (2 +- 6.324)/6$

$-> x = 8.324/6$ $or$ $x= - 4.324/6$

$:.$ $X = 1.387 , -0.720$

Answer 2 · 2017-06-28T11:38:46

Use the $B^2 - 4AC = D$

then $x1 ; x2 = (- B +- sqrt(D))/(2A)$

When $A= 3, B = -2, C = -3$

$D = 4 + 36$

$D = 40$

$x1 = (2+sqrt(40))/6$

$x2 = (2-sqrt(40))/6$

How do you solve 0=3x^2-2x-3 0=3x2−2x−30=3x^2-2x-3 using the quadratic formula?