How do you solve $x^{2} + x - 10 = 0$ $x^2+x-10=0$ ?

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Answer 1 · 2016-02-26T16:17:17

The solutions are:
$x = \frac{- 1 + \sqrt{41}}{2}$ $x= (-1+sqrt(41))/2$

$x = \frac{- 1 - \sqrt{41}}{2}$ $x= (-1-sqrt(41))/2$

$x^{2} + x - 10 = 0$ $x^2 +x -10=0$

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:
$a = 1, b = 1, c = - 10$ $a=1, b=1, c=-10$

The Discriminant is given by:

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

$= (1)^2-(4* 1*-10)$

$= 1 +40 = 41$

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

$x = ((-1)+-sqrt(41))/(2*1) = (-1+-sqrt(41))/2$

The solutions are:
$x= (-1+sqrt(41))/2$

$x= (-1-sqrt(41))/2$

How do you solve x^2+x-10=0x2+x−10=0x^2+x-10=0?