Posted by **Mary** on Saturday, June 27, 2009 at 10:56am.

How do i caluclate the roots of the equation for:

x^2-3x-12=0

(x^2-4)=0 (x+3)=0

x^2=4 x+3+0

x=4 x=-3

Am i on the right path or just way off??

- mathematics -
**Reiny**, Saturday, June 27, 2009 at 11:15am
I am assuming there is only the one equation

x^2 - 3x - 12 = 0, because what follows makes no sense

First of all your quadratic does not factor, so you will have to either use the quadratic equation formula or complete the square.

let's use the formula:

x = (3 ± √(9-4(1)(-12))/2

= (3 ± √57)/2

Use your calculator if you need a decimal equivalent.

- Mathematics -
**Mary**, Saturday, June 27, 2009 at 11:54am
Ok i may be dense but i don't ge tthe formula cause when i work out

3 } ã(9-4(1)(-12))/2

I get 60 not 57, and i don't understand how you got 9 or 4 either

- mathematics -
**Reiny**, Saturday, June 27, 2009 at 12:03pm
for any quadratic equation

ax^2 + bx + c = 0

x = (-b ± √(b^2 - 4ac))/(2a)

so in our case

a = 1, b = -3, and c = -12

look at the second example on this page

http://www.nipissingu.ca/calculus/tutorials/quadratics.html

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