anonymous
  • anonymous
Which expression gives the solutions of –5 + 2x² = –6x?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
add 6x to both sides
myininaya
  • myininaya
\[2x^2+6x-5=0\]
myininaya
  • myininaya
Now use the quadratic formula! :)

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More answers

myininaya
  • myininaya
a=2,b=6,c=-5
anonymous
  • anonymous
whats the quadratic formula?
myininaya
  • myininaya
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
i dont know how to use that. D:
myininaya
  • myininaya
Do you know how to complete the square?
anonymous
  • anonymous
no. :/ im horrible at math.
myininaya
  • myininaya
hmmm... so what method do you want to use?
anonymous
  • anonymous
i dont know any methods to solve that.
ash2326
  • ash2326
@Qudrex I'll help you.
anonymous
  • anonymous
okay.
ash2326
  • 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
  • ash2326
This formula gives us two values of x which satisfies the quadratic equation
ash2326
  • ash2326
Did you understand till now?
anonymous
  • anonymous
formulas confuse me.
ash2326
  • ash2326
what's confusing you?
anonymous
  • anonymous
the formula i dont know how to use it.
ash2326
  • 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
  • anonymous
yes. i think i understand that.
ash2326
  • 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
  • anonymous
kinda
ash2326
  • 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
  • anonymous
yes
ash2326
  • 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
  • anonymous
yeah.
ash2326
  • ash2326
So whenever you've to find roots of quadratic equation, use this formula:D
anonymous
  • anonymous
okay!

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