SOLUTION: I don't understand how to continue this problem. The problem is: 10x^2+10=8-6x I know that I have to subtract the 8 then the -6x to get 0 I got: 10x^2+6x+2=0 From there, would

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I don't understand how to continue this problem. The problem is: 10x^2+10=8-6x I know that I have to subtract the 8 then the -6x to get 0 I got: 10x^2+6x+2=0 From there, would      Log On


   

Question 729504: I don't understand how to continue this problem. The problem is:
10x^2+10=8-6x
I know that I have to subtract the 8 then the -6x to get 0
I got:
10x^2+6x+2=0
From there, would I have to do the quadratic formula? Because the number you have to square root it by is a negative number and we haven't learned that yet. I tried factoring but I got:
5x^2+3x+1=0
What do I do from here?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
you did good job, now just use quadratic formula to solve for x
5x%5E2%2B3x%2B1=0..note that coefficient a=5, b=3 and c=1
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+...plug coefficients in
x+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A5%2A1+%29%29%2F%282%2A5%29+

x+=+%28-3+%2B-+sqrt%28+9-20%29%29%2F10+
x+=+%28-3+%2B-+sqrt%28+-11%29%29%2F10+......you have sqrt of negative number and that means there is no real solutions only imaginary and graph will not intersect with x-axis, and solutions are complex numbers
x+=+%28-3+%2B-+sqrt%28+-11%29%29%2F10+......write sqrt%28-11%29 as sqrt%28-1%2A11%29

x+=+%28-3+%2B-+sqrt%28+-1%2A11%29%29%2F10+...since sqrt%28-1%29=i we have

x+=+%28-3+%2B-+sqrt%2811%29i%29%2F10+

x+=+%28-3+%2B-+3.32i%29%2F10+
solutions:
x+=+%28-3+%2B+3.32i%29%2F10+
x+=+-3%2F10+%2B+3.32i%2F10+
or

x+=+%28-3+-+3.32i%29%2F10+
x+=+-3%2F10+-3.32i%2F10+

+graph%28+600%2C+600%2C+-5%2C5%2C+-5%2C+10%2C5x%5E2%2B3x%2B1%29+