How do you solve $2 x^{2} + x + 7 = 0$ $2x^2 + x + 7 = 0$ ?

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Answer 1 · 2016-03-09T14:10:05

The solutions are:
$x = \frac{- 1 + \sqrt{- 55}}{4}$ $x=(-1+sqrt(-55))/4$

$x = \frac{- 1 - \sqrt{- 55}}{4}$ $x=(-1-sqrt(-55))/4$

$2 x^{2} + x + 7 = 0$ $2x^2 + x+7=0$

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:
$a = 2, b = 1, c = 7$ $a=2, b=1, c=7$

The Discriminant is given by:

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

$= (1)^2-(4*2*7)$

$= 1 - 56 = - 55$

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-1)+-sqrt(-55))/(2*2) = (-1+-sqrt(-55))/4$

The solutions are:
$x=(-1+sqrt(-55))/4$

$x=(-1-sqrt(-55))/4$

How do you solve 2x^2 + x + 7 = 02x2+x+7=02x^2 + x + 7 = 0?