I hate math. Think its mainly because i've been making silly mistakes, but here it is: 3X^2+14X-5=0
hides in corner
anyone care to help our give guidance? :)

ax^2+bx+c=0

(-b + √ (b^2 - 4ac))/2a


for the equation X^2+14X-5=0
a = 3
b = 14
c = -5

so substitute the values in:

[-14 + √ (14^2 - 4(3)(-5) ] /2(3)

[-14 +√( 196 + 60)] /6

[-14+√(256)] /6

and because the quadratic equation is +, you need to use the equation once with addition, and once with subtraction

[-14+√(256)] /6 = -.333

[-14 - √(256)] /6 = -5

Thank you!

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