Quadratic Formula: 5 x^2 +x=5

is it just -1 +- the square root of 101/10?

5x^2+x-5=0 ax^2+bx+c=0 then \[x=-b \pm \sqrt{b^2 -4ac} \div \]

Rewrite: \[5x^2+x-5=0\] Quadratic formula: \[x=(-b \pm \sqrt{b^2-4ac)})/2\] So x = (1/2)(-1 +/- sqrt(1-4times1times(-5)) = (1/2)(-1+/- sqrt21)

Yeah i get that but what after that??

div 2a

Yes i know thats too far into decimals to work with

x=-1/2 +/- (1/2)sqrt21=-0.5+/- 2.29 (from calculator) x= -2.79 or +1.79

yes right

Something's wrong. These solutions don't satisfy the original equation. Back soon.

I forgot to divide by 2a = 10

I know thats what im saying

So sorry - carelessness on my part. Here we go again: \[x=(-1\pm \sqrt{1-4\times5\times(-5)})/10\] \[x=(-1\pm \sqrt{101})\div10\] x = (-1 +/- 10.0499)/10 x = -1.10499 or 0.90499

Still not right!

guyc, I got the same answer: \[\frac{-1 \pm \sqrt{101}}{10}\] Now the square root of 101 does not simplify since 101 is prime, so this is a final answer.

Yes, I agree with cruffo. Incidentally, my decimal answers do in fact satisfy the original equation. I do hope I didn't confuse you with my ramblings!

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