Question 33217
Hey check this out in case I do it wrong myself!
 
You know the quadratic formula is {{{X = (-B +- sqrt( B^2-4*A*C ))/(2*A)}}} .
You also know that A, B, and C come in the equation like this Ax^2+Bx+C.

In this case, A=3 B=-4 C=-5. You plug them in to the equation (replace letters with numbers) and we have {{{X=  (-(-4) +- sqrt( (-4)^2-4*(3)*(-5) ))/(2*(3))}}}  .

Well your answer breaks down to {{{(4 +- sqrt( 16+60))/6}}} .
[the four is positive because 2 negatives make a positive, the 16 is -4 squared, it's + 60 because -4*3*-5 = POSITIVE 60, and the 6 is 2*3.]

Now, we simplify the square root. {{{sqrt(76)}}} is not an integer (whole number), so we simplify it. 

76 is evenly divisible by 2, because its 1's place digit is evens. 76/2 is 38, so 38(2)=76.  Still, 38 ends evenly, so it is also divisible by 2.  38/2=19.  so 19(2)=38.  Nineteen is a prime number, not evenly divisible by any whole numbers but 1 and itself.  19*2*2=76.

We have 2*2*19 as our primes. 2*2 is the same as {{{2^2}}} is the same as 4.  {{{sqrt(4)}}} = 2, so we take 4's root (2) outside of the square root.

Our answer is {{{(4 +- 2*sqrt(19))/6}}}.  One plus or minus can go, since either way we will plus or minus, and the 4,2, and 6 have a common factor of 2.  The equation simplifies again to {{{(2 +- 1*sqrt(19))/3}}}.  We don't need the one: {{{(2 +- sqrt(19))/3}}}.   The answers are {{{(2+sqrt(19))/3}}} and {{{(2-sqrt(19))/3}}}.

I'll allow your imagination to take over the second problem, the answers are 1/2 and 1. IM me if you need more help.