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

Question

myininaya · Answer

add 6x to both sides

myininaya · Answer

$$2x^2+6x-5=0$$

myininaya · Answer

Now use the quadratic formula! :)

myininaya · Answer

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

anonymous · Answer

whats the  quadratic formula?

myininaya · Answer

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

anonymous · Answer

i dont know how to use that. D:

myininaya · Answer

Do you know how to complete the square?

anonymous · Answer

no. :/ im horrible at math.

myininaya · Answer

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

anonymous · Answer

i dont know any methods to solve that.

ash2326 · Answer

@Qudrex I'll help you.

anonymous · Answer

okay.

ash2326 · Answer

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 · Answer

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

ash2326 · Answer

Did you understand till now?

anonymous · Answer

formulas confuse me.

ash2326 · Answer

what's confusing you?

anonymous · Answer

the formula i dont know how to use it.

ash2326 · Answer

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 · Answer

yes. i think i understand that.

ash2326 · Answer

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 · Answer

kinda

ash2326 · Answer

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 · Answer

yes

ash2326 · Answer

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 · Answer

yeah.

ash2326 · Answer

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

anonymous · Answer

okay!