need the solutions please step by steps.
4x^2+x-1=0

Question

anonymous · Answer

Do Quadratic equation - A=4 - B=1 - C= -1

$$(-1\pm \sqrt{1-4\times4\times -1})\div 2\times4$$

Do that equation for the plus and the minus and you'll have two answers

anonymous · Answer

I really don't understand

anonymous · Answer

It's a formulae to solve any quadratic equation of the form A*x^2+B*x+C = 0.

anonymous · Answer

The formulae is:

$$( -B \pm \sqrt{B ^{2}-4AC} ) / 2A$$

You identify A, B and C and you use the formulae above to get two answers. If one of you answers turns out to be an imaginary number, you discard it.

anonymous · Answer

So, essentially what you're trying to do is solve for what the possible value(s) of "x" might be.  when you have an equation like this, the number next to x^2 is called "A", the number next to x is called "B" and the number all by itself (called a constant) is called "C."  so, for this equation, which looks like $$Ax ^{2}+Bx+C$$ YOUR "A" would equal 4, your "B" would equal 1 (because no number next to a variable like "x" implies a value of 1) and your "C" would equal -1 (since the value of "C" is being SUBTRACTED from the rest of the equation in your problem).  

Now that we have values for A, B and C, we can use them in this equation called the QUADRATIC FORMULA.  this formula looks like this:
$$x=(-B \pm \sqrt{B ^{2}-4AC})/2A$$

Note that the 2A is dividing EVERYTHING on top of it (that is, everything in parentheses).  

Now, after you plug in your values for each of the letters, you get 
$$x=(-1\pm \sqrt{1^{2}-4*(4)(-1)})/(2*4)$$

which simplifies into 
$$x= (-1\pm \sqrt{17})/8$$

This gives you two different answers, one for when you ADD the sqrt of 17 in the equation, and one for when you SUBTRACT it from the equation.  

the two equations are 

$$x=(-1+\sqrt{17})/8=.39$$and$$x= (-1-\sqrt{17})/8=-.64$$
Therefore, your two values for x are, respectively, .39 and -.64.

If you have any more questions feel free to send me a message!

anonymous · Answer

note that, for the first equation, it should actually be 

$$Ax ^{2}+Bx+C=0$$

sorry about that!

anonymous · Answer

thats ok thanks.