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?
+26396 \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} $}}
+26396 \small{\text{$ \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} $}}
