# I know this is two different questions but i thought id save a question and ask both in one?

Question:how would i solve something with two different x's in it like this math equation?

2x^2 - 7x= -3

and the second one is how can i tell if there is a solution, 2, or none on a problem like 3z^2 + 6z = -3

thankyou

Answers:
2x^2 - 7x= -3
2x^3 -7x +3=0, use quadratic eq:
x= ( 7+- sqrt( 49 -4(2)(3)) )/4
= ( 7 +- 5)/4 then
x = 12/4=3 or x = 2/4= 1/2

3z^2 + 6z = -3
3z^2 + 6z +3 =0
z^3 + 2z + 1 = ( z+1)^2 =0 => there is only 1 solution z = -1, but
you can also use the quadratic formula , if the discriminant b^2 -4ac>0 then there are 2 different solutions, if b^2 -4ac =0 then there is 1 solution
and if b^2 -4ac <0 then there are no solutions at all
OK, I could be wrong here, but I believe they ARE solved. You have three distinct "numbers" so they cannot be worked out any further.

For instance, you can't add x and x^2 together and come up with 2x^2. You CAN add 2x^2 + 3x^2 to get 5x^2 however.

So, you can see there is nothing in your equations that you can add together.

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