Question

Need help
2x^2+10x-14=0

Answer 1

I tried 2(x^2+5x-7)=0

Answer 2

How are you trying to solve this? are we using the quadratic formula, are we solving for x, are we factoring? @silvernight269

Answer 3

Solving for x

Answer 4

Okay. Honestly, this can be done with the quadratic formula

Answer 5

Just for review, it looks like this: \[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

Answer 6

To make it equal to 0?

Answer 7

just one moment ? x_1,2 = this formule bc. there are + and - so result two roots

Answer 8

there are, but the roots are not exact

Answer 9

\(ax^2 + bx = c\)

so we have to find the a, b, and c terms and then we can plug those in and solve for the various x's

Answer 10

So this means that a would equal 2, b =10, and c would be 14 since when you bring it to the other side, it becomes positive 14.

Answer 11

It still doesn’t make sense

Answer 12

okay. Let's try plugging everything in then

Answer 13

\(x_{(1,2)} = \frac{ -10 \pm \sqrt{(10)^2 - 4(2)(14)} }{ 2(2) }\)

Answer 14

It would actually look more along the lines of this: x_{1,2} = \frac{ -10 \pm \sqrt{ 212 } }{ 4 }

Answer 15

oh, oops. Let's try that again: \(x_{1,2} = \frac{ -10 \pm \sqrt{ 212 } }{ 4 }\)

Answer 16

Why 212

Answer 17

because I'm realizing that I goofed in telling you that c = 14. It's actually -14. The reason for this is because the equation you gave was already in quadratic form and I didn't realize. It's the negative that changes the answer so drastically

\(10^2 - 4 (2) (-14)\) = 212

Answer 18

Oooooooh

Answer 19

\(x_1 = \frac{ -10~+~\sqrt{ 212 } }{ 4 } = -\frac{5}{2}+\frac{1}{2}\sqrt{53}\)
\(x_2 = \frac{ - 10~-~\sqrt{ 212 } }{ 4} = -\frac{5}{2} - \frac{1}{2}\sqrt{53}\)

Answer 20

This is what things should look like once everything is simplified, including the radicals. Does this make sense?

Answer 21

Kinda

Answer 22

What do you still have questions on?

Answer 23

Like, you know, where in the process?

Answer 24

Actually I think I’m ok

Answer 25

you sure?

Answer 26

Yea

Answer 27

Okay, glad to have helped!