SOLUTION: Solve 2x^2 + 5x - 1 = 0 by using quadratic formula. Use coefficients, a=2, b=5, c=-1. My Atttempt: Using quadratic formula w/coefficients given -(5)+/- sqrt (5^2 - 4(2)

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Question 177509: Solve 2x^2 + 5x - 1 = 0 by using quadratic formula. Use coefficients, a=2, b=5, c=-1. My Atttempt: Using quadratic formula w/coefficients given

-(5)+/- sqrt (5^2 - 4(2)(-1)
x= _______________________________
2(2)

-(5)+/- sqrt (25-4(2)(-1)
x=_____________________________
2(2)

-5 +/- sqrt (25+8)
x=______________________
4

-5 +/- sqrt (33)
x=__________________
4
I got stuck here!!



Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

2x%5E2%2B5x-1=0 Start with the given equation.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=2, b=5, and c=-1


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%285%29+%2B-+sqrt%28+%285%29%5E2-4%282%29%28-1%29+%29%29%2F%282%282%29%29 Plug in a=2, b=5, and c=-1


x+=+%28-5+%2B-+sqrt%28+25-4%282%29%28-1%29+%29%29%2F%282%282%29%29 Square 5 to get 25.


x+=+%28-5+%2B-+sqrt%28+25--8+%29%29%2F%282%282%29%29 Multiply 4%282%29%28-1%29 to get -8


x+=+%28-5+%2B-+sqrt%28+25%2B8+%29%29%2F%282%282%29%29 Rewrite sqrt%2825--8%29 as sqrt%2825%2B8%29


x+=+%28-5+%2B-+sqrt%28+33+%29%29%2F%282%282%29%29 Add 25 to 8 to get 33


x+=+%28-5+%2B-+sqrt%28+33+%29%29%2F%284%29 Multiply 2 and 2 to get 4.


x+=+%28-5%2Bsqrt%2833%29%29%2F%284%29 or x+=+%28-5-sqrt%2833%29%29%2F%284%29 Break up the expression.


So the answers are x+=+%28-5%2Bsqrt%2833%29%29%2F%284%29 or x+=+%28-5-sqrt%2833%29%29%2F%284%29


which approximate to x=0.186 or x=-2.686