SOLUTION: I wrote problem two ways: I think this one however, is the correct one f(x)=3x^2 + x - 5 f(x)=3x^2 + x - 5 3x^2 + x -5 = 0

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I wrote problem two ways: I think this one however, is the correct one f(x)=3x^2 + x - 5 f(x)=3x^2 + x - 5 3x^2 + x -5 = 0       Log On


   

Question 246109: I wrote problem two ways: I think this one however, is the correct one
f(x)=3x^2 + x - 5
f(x)=3x^2 + x - 5
3x^2 + x -5 = 0
Answer by marcsam823(57) About Me  (Show Source):
You can put this solution on YOUR website!
Since there is no easy way to factor this equation, use the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
the general formula for a quadratic is
ax%5E2+%2B+bx+%2B+c+=+0

Your equation is:
3x%5E2+%2B+x+-5+=+0

Therefore:
a=3
b=1
c=-5

Substitute and solve for x:
x+=+%28-%281%29%2B-+sqrt%28+1%5E2-4%2A3%2A-5%29%29%2F%282%2A3%29+

x+=+%28-%281%29%2B-+sqrt%28+1-%28-60%29%29%29%2F6+

x+=+%28-%281%29%2B-+sqrt%28+1%2B60%29%29%2F6+

x+=+%28-1+%2B-+sqrt%28+61%29%29%2F6+

There are 2 answers: (this are the x -coordinates of the points where the parabola crosses the x-axis):
%28-1+%2B+sqrt%2861%29%29%2F6+=+1.135
%28-1+-+sqrt%2861%29%29%2F6+=+1.468