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

Originally Posted by ceedee

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

\begin{align*}5x^2-3x+12&=3(x+4) \\&=3x+12\\5x^2-6x&=0\\\text{Now you solve} \end{align*}

Originally Posted by ceedee
$5x^2 - 6x = 0$

$x(5x-6) = 0$

can you finish it now?

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

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

$5x^2 - 6x = 0$

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

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

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 $x=\dfrac 6 5$