How do you solve $x^{2} + x - 12 = 0$ $x^2 + x - 12 = 0$ ?

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How do you solve $x^{2} + x - 12 = 0$ $x^2 + x - 12 = 0$ ?

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Answer 1 · 2016-03-09T14:04:46

The solutions are:

$x = 3$ $x =3$

$x = - 4$ $x = -4$

Explanation:

$x^{2} + x - 12 = 0$ $x^2 + x-12 =0$

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

$a = 1, b = 1, c = - 12$ $a=1, b=1, c=-12$

The Discriminant is given by:

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

$= (1)^2-(4*1*(-12))$

$= 1 +48 =49$

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

$x = ((-1)+-sqrt(49))/(2*1) = (-1+-7)/2$

$x = (-1+7)/2 = 6/2 =3$

$x = (-1-7)/2 = -8/2 = -4$

The solutions are:

$x =3$

$x = -4$

Answer 2 · 2016-03-09T14:17:35

$x=-4,3$

Explanation:

$x^2+x-12=0$

This is in the form of a Quadratic equation
(in form $ax^2+bx+c=0$ )

Where

$a=1,b=1,c=-12$

Use Quadratic formula

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Substitute the values

$rarrx=(-1+-sqrt(1^2-4(1)(-12)))/(2(1))$

$rarrx=(-1+-sqrt(1-(-48)))/(2)$

$rarrx=(-1+-sqrt(1+48))/(2)$

$rarrx=(-1+-sqrt49)/(2)$

$rarrx=(-1+-7)/2$

Now we have two values for $2$

$x=(-1-7)/2,x=(-1+7)/2$

If we solve for it,We get

$rArrx=-4,3$

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How do you solve $x^{2} + x - 12 = 0$ $x^2 + x - 12 = 0$ ?

2 Answers

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