Question

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??

Answer 1

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.

Answer 2

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

Answer 3

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