# Thread: Quadratic Equation

1. ## Quadratic Equation

5x2
-3x+12=3(x+4)

There's supposed to be two different possible solutions for X, please help me find them.

2. ## Re: Quadratic Equation Originally Posted by ceedee 5x2
-3x+12=3(x+4)

There's supposed to be two different possible solutions for X, please help me find them.
\begin{align*}5x^2-3x+12&=3(x+4) \\&=3x+12\\5x^2-6x&=0\\\text{Now you solve} \end{align*}

3. ## Re: Quadratic Equation

Yes I also get that far, but I don't get further...help please and can you explain your answer then also please?

4. ## Re: Quadratic Equation Originally Posted by ceedee Yes I also get that far, but I don't get further...help please and can you explain your answer then also please?
$5x^2 - 6x = 0$

$x(5x-6) = 0$

can you finish it now?

If the product of two numbers is zero what can you say about at least one of those numbers?

5. ## Re: Quadratic Equation

Thank you for the answer, is it X=2 ?

6. ## Re: Quadratic Equation

$5x^2 - 6x = 0$

$\large ax^2+bx+c=0\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$a=5,~b=-6,~\&~c=0$

7. ## Re: Quadratic Equation Originally Posted by ceedee Thank you for the answer, is it X=2 ?
no...

$x(5x-6) = 0$

in order for the product to equal zero at least one of the factors must equal zero.

So either

$x=0$

or

$5x-6=0 \rightarrow x = \dfrac 6 5$

so $x = 0$ OR $x=\dfrac 6 5$

8. ## Re: Quadratic Equation

Thank you very much

Thank you!!