Question

Solve the equation.
2x^2-5x=3

Answer 1

-3x=3
x=-1 is your answer

Answer 2

I think :P

Answer 3

it can't be just substitute if you substitute -1 then the answer is 7 but we are trying to get 3

Answer 4

It's a rational expression. You don't just plug the numbers in.

Answer 5

Ok umm but the answer still has to be a negative right???

Answer 6

I'm not sure. It will be two numbers. I think it is 3,5

Answer 7

I think it is just 3

Answer 8

It can't be just 3. lol

Answer 9

why?

Answer 10

are there answer choices?

Answer 11

\[2x^2-5x=3\rightarrow 2x^2-5x-3=0\]Then you use the quadratic formula.\[-(-5)\pm \frac{ \sqrt{(-5)^2-4(2)(-3)} }{ 2(2) }\]Can you solve from there?

Answer 12

I am confused because I haven't been working them out this way. Not quite sure where to start?

Answer 13

Oops, everything should be over 2(2), including the -(-5) part

Answer 14

How have you been working them out? Have you used the quadratic formula for these at all?

Answer 15

I am confused too, I haven't learned any of that

Answer 16

Yes I have been using the quadratic formula but I somehow was finding the right answer using a different method. I have no clue what I was doing. Whatever I was doing isn't working on this.

Answer 17

Were you factoring things out, and setting them to zero? For example
x^2+9x+18 -->(x+3)(x+6)
then
x+3=0 --> x=-3 and
x+6=0-->x=-6 ?
Sometimes you can't factor the quadratic out so you need to use the quadratic formula to get your zeros.

Answer 18

That is what I have been doing. This can't be factored though?

Answer 19

No this one can't be factored. In this case you would need to utilize the quadratic formula.

Answer 20

Ok so with the formula you worked out above where would I start?

-5=-25+12/4

Answer 21

Alright so we have your quadratic in standard form:
\[ax^2+bx+c=0\]
\[2x^2-5x-3=0\]
We plug your corresponding numbers into our quadratic formula:
\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\rightarrow \frac{ -(-5)\pm \sqrt{(-5)^2-4(2)(-3)} }{ 2(2) }\]

Answer 22

We split the equation into a plus and minus version to get
\[\frac{ 5+\sqrt{49} }{ 4 } and \frac{ 5-\sqrt{49} }{ 4 }\]

Answer 23

2x^2-5x - 3 =0
a = 2 b = - 5 c =- 3
Using the magical quadratic formula.
\[\large x = \frac{ 5 \pm \sqrt{25 - 4(2)(-3)} }{ 4 } \]
\[\large x = \frac{ 5 \pm \sqrt{49} }{ 4 } = \frac{ 5 \pm 7 }{ 2 }\]
\[\large x = \frac{ 5 + 7 }{ 2 } = 6\]
\[\large x =\frac{ 5-7 }{ 2 } = -1\]

Answer 24

So we have that \[x=\frac{ 5+7 }{ 4 }=3 ; x=\frac{ 5-7 }{ 4 }=-\frac{ 1 }{ 2 }\]

Answer 25

Ok. That is very confusing! Thank you for your help!!

Answer 26

Shamil98, you forgot to continue your 4 in your denominator, you flipped it back to 2.