Solve $x$ in $12x^2 + 30x = 0$ I need to solve $x$ in $12x^2 + 30x = 0$.
I started by dividing everything by $12$.
$$x^2 + 2.5x = 0$$
I then used $pq$.
$$x = 1.25 \pm \sqrt{1.5625}$$
$$X_1 = 1.25 + 1.25 = 2.5$$
$$X_2 = 1.25 - 1.25 = 0$$
But when i look at the key, it says the answer should be
$$X_1 = -2.5$$
$$X_2 = 0$$
I don't know what i did wrong, so any help is appreciated.
 A: I believe your problem is that is should be
$$x=-1.25\pm\sqrt{1.5625}.$$
Notice the minus in front of $1.25$.
A: Notice that $12x^2+30x=0$ is equivalent to $x(12x+30)=0$
Thus $x=0$ or $12x+30=0$. Therefore, $x=0$ or $x=-\frac{30}{12}=-2.5$
A: You should be very careful using signs.
$$x^2+2.5x=0$$ or, $$x(x+2.5)=0$$ or, $$X_1=0$$ and $$X_2=-2.5$$
A: A simpler way to proceed:
$$
12 x^2 + 30 = x(12x+30),
$$
so the roots are:


*

*the root of $x$, namely, $x = 0$, and

*the root of $12x + 30$, or $x = -30/12 = -15/6 = -2.5$.

A: if you want to use the quadratic formula, the solution is 
$$12x^2+30x=0$$
$$x^2+2.5x+0=0$$
$$x=-1.25\pm\sqrt{1.25^2-0}$$
A: Notice, factorize the quadratic polynomial as follows $$12x^2+30x=0$$ $$6x(2x+5)=0$$
$$\implies 5x=0\iff \color{red}{x=0}$$
or $$\implies 2x+5=0\iff \color{red}{x}=-\frac{5}{2}=\color{red}{-2.5}$$
A: Suppose you have a general quadratic equation. Then
$$\eqalign{
  & a{x^2} + bx + c = 0  \cr 
  & \Delta  = {b^2} - 4ac  \cr 
  & if\,\,\Delta  \ge 0\,\,then\,\,{x_{1,2}} = {{ - b \pm \sqrt \Delta  } \over {2a}}\,\,\,else\,\,there\,\,is\,\,no\,\,real\,\,root. \cr} $$
Hence, in your case
$$\left\{ \matrix{
  a = 1 \hfill \cr 
  b = 2.5 \hfill \cr 
  c = 0 \hfill \cr}  \right.\,\,\, \to \,\,\,\Delta  = \sqrt {{{\left( {2.5} \right)}^2}}  = \left| {2.5} \right|\,\,\, \to \,\,\,{x_{1,2}} = {{ - 2.5 \pm \left| {2.5} \right|} \over 2}\,\,\,\, \to \,\,\,\left\{ \matrix{
  {x_1} = 0 \hfill \cr 
  {x_2} =  - 2.5 \hfill \cr}  \right.$$
