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

Mathematics
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add 6x to both sides
\[2x^2+6x-5=0\]
Now use the quadratic formula! :)

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Other answers:

a=2,b=6,c=-5
whats the quadratic formula?
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
i dont know how to use that. D:
Do you know how to complete the square?
no. :/ im horrible at math.
hmmm... so what method do you want to use?
i dont know any methods to solve that.
@Qudrex I'll help you.
okay.
A standard quadratic equation is given as \[ax^2+bx+c=0\] its roots or its solution is given by the following formula \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
This formula gives us two values of x which satisfies the quadratic equation
Did you understand till now?
formulas confuse me.
what's confusing you?
the formula i dont know how to use it.
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?
yes. i think i understand that.
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?
kinda
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?
yes
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?
yeah.
So whenever you've to find roots of quadratic equation, use this formula:D
okay!

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