40x+10x^2=o ? what i did was divide it by 10 so it became 4x + x^2=0
Then moved 4x to the otherside so it becamse x^2 = -4x then the square root of that? Is this a solution or am i doing wrong?
+26367 $$\small{\text{$
\begin{array}{rcl}
40x+10x^2 &=& 0 \\\\
\underbrace{ \textcolor[rgb]{150,0,0}{\mathbf{10x}} }_{=0}\cdot ( \underbrace{ \textcolor[rgb]{0,0,150}{\mathbf{4+x}} }_{=0} ) &=& \mathbf{ 0 } \\\\
\end{array}
$}}\\\\
\textcolor[rgb]{150,0,0}{\mathbf{10x}} = 0 \qquad | \qquad : 10 \\
\boxed{ \textcolor[rgb]{150,0,0}{\mathbf{x}} = 0}\\\\
\textcolor[rgb]{0,0,150}{\mathbf{4+x}} = 0 \qquad | \qquad -4 \\
\boxed{ \textcolor[rgb]{0,0,150}{\mathbf{x}} = -4}$$
$$\\\small{\text{check: }}\\
\small{\text{
$
\begin{array}{rcl}
40\cdot \textcolor[rgb]{150,0,0}{0} + 10\cdot \textcolor[rgb]{150,0,0}{0}^2 &=& 0 \\
0+0 &=& 0 \small{\text{ okay}}\\
\end{array}
$}}\\\\
\small{\text{
$
\begin{array}{rcl}
40\cdot (\textcolor[rgb]{0,0,150}{-4}) + 10\cdot (\textcolor[rgb]{0,0,150}{-4})^2 &=& 0 \\
-160 + 10\cdot 16 &=& 0\\
-160 + 160 &=& 0 \\
0 &=& 0 \small{\text{ okay}}
\end{array}
$}}$$
+26367 $$\small{\text{$
\begin{array}{rcl}
40x+10x^2 &=& 0 \\\\
\underbrace{ \textcolor[rgb]{150,0,0}{\mathbf{10x}} }_{=0}\cdot ( \underbrace{ \textcolor[rgb]{0,0,150}{\mathbf{4+x}} }_{=0} ) &=& \mathbf{ 0 } \\\\
\end{array}
$}}\\\\
\textcolor[rgb]{150,0,0}{\mathbf{10x}} = 0 \qquad | \qquad : 10 \\
\boxed{ \textcolor[rgb]{150,0,0}{\mathbf{x}} = 0}\\\\
\textcolor[rgb]{0,0,150}{\mathbf{4+x}} = 0 \qquad | \qquad -4 \\
\boxed{ \textcolor[rgb]{0,0,150}{\mathbf{x}} = -4}$$
$$\\\small{\text{check: }}\\
\small{\text{
$
\begin{array}{rcl}
40\cdot \textcolor[rgb]{150,0,0}{0} + 10\cdot \textcolor[rgb]{150,0,0}{0}^2 &=& 0 \\
0+0 &=& 0 \small{\text{ okay}}\\
\end{array}
$}}\\\\
\small{\text{
$
\begin{array}{rcl}
40\cdot (\textcolor[rgb]{0,0,150}{-4}) + 10\cdot (\textcolor[rgb]{0,0,150}{-4})^2 &=& 0 \\
-160 + 10\cdot 16 &=& 0\\
-160 + 160 &=& 0 \\
0 &=& 0 \small{\text{ okay}}
\end{array}
$}}$$
