- anonymous

Which expression gives the solutions of –5 + 2x² = –6x?

- chestercat

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- myininaya

add 6x to both sides

- myininaya

\[2x^2+6x-5=0\]

- myininaya

Now use the quadratic formula! :)

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## More answers

- myininaya

a=2,b=6,c=-5

- anonymous

whats the quadratic formula?

- myininaya

\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

- anonymous

i dont know how to use that. D:

- myininaya

Do you know how to complete the square?

- anonymous

no. :/ im horrible at math.

- myininaya

hmmm... so what method do you want to use?

- anonymous

i dont know any methods to solve that.

- ash2326

@Qudrex I'll help you.

- anonymous

okay.

- ash2326

A standard quadratic equation is given as
\[ax^2+bx+c=0\]
its roots or its solution is given by the following formula
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

- ash2326

This formula gives us two values of x which satisfies the quadratic equation

- ash2326

Did you understand till now?

- anonymous

formulas confuse me.

- ash2326

what's confusing you?

- anonymous

the formula i dont know how to use it.

- ash2326

It's very easy, I'll show you
your question is
-5+2x^2=6x
The first step is to bring all the terms to one side
\[-5+2x^2=6x\]
Let's subtract both sides by 6x
we get
\[-5+2x^2-6x=6x-6x\]
we'll get
\[-5+2x^2-6x=0\]
Now let's write this in descending powers of x
\[2x^2-6x-5\]
Did you understand till now?

- anonymous

yes. i think i understand that.

- ash2326

Great now let's compare this with standard equation
\[ax^2+bx+c=0\]
\[2x^2-6x+5=0\]
we have
\[a=2,\ b=-6\ and\ c=5 \]
Did you get this?

- anonymous

kinda

- ash2326

Now its solution is given by
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
Let's substitute a=2, b=-6 and c=-5
\[x=\frac{-(-6)\pm \sqrt{(-6)^2-4\times (2)\times (-5)}}{2\times (2)}\]
We get now
\[x=\frac{6\pm \sqrt{36-(-40)}}{4}\]
\[x=\frac{6\pm \sqrt{36+40}}{4}\]
\[x=\frac{6\pm \sqrt{76}}{4}\]
Did you understand this?

- anonymous

yes

- ash2326

We've almost found the solution, we could simplify it more
We have
\[x=\frac{6\pm \sqrt{76}}{4}\]
\[\sqrt {76}= \sqrt{4\times 19}=2\sqrt {19}\]
so
\[x=\frac{6\pm 2\sqrt{19}}{4}\]
divide the whole equation by 2
we get
\[x=\frac{3\pm \sqrt {19}}{2}\]
so the two solutions\roots are
\[x=\frac{3+ \sqrt {19}}{2}, \frac{3- \sqrt {19}}{2}\]\]
Did you understand?

- anonymous

yeah.

- ash2326

So whenever you've to find roots of quadratic equation, use this formula:D

- anonymous

okay!

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