How do you solve $3 x^{2} + 5 x + 2 = 0$ $3x^2 + 5x + 2 = 0$ ?

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Answer 1 · 2016-05-29T16:16:39

The solutions are:

$x = - \frac{2}{3}$ $color(blue)(x = -2/3$

$x = - 1$ $color(blue)(x = -1$

$3 x^{2} + 5 x + 2 = 0$ $3x^2 +5x + 2 = 0$

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:

$a = 3, b = 5, c = 2$ $a=3, b=5, c=2$

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (5)^2-(4 * 3 * 2)$

$= 25 - 24 = 1$

The solutions are found using the formula:

$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-5)+-sqrt(1))/(2*3) = (-5 +- 1 )/8$

$x = ( -5 + 1 ) / 6 = -4/6 = -2/3$

$x = ( -5 - 1 ) / 6 = -6/6 = -1$

How do you solve 3x^2 + 5x + 2 = 03x2+5x+2=0 3x^2 + 5x + 2 = 0?