Andregradslikninger

$\begin{array}{rcl}2x\left(x-1\right)& = & 0\\ & & \\ 2x& =& 0 \mathrm{eller} x-1=0\\ & & \\ x_1& =& 0 \mathrm{eller} x_2=1\end{array}$

$\begin{array}{rcl}12x-3x^2& = & 0\\ & & \\ 3x\left(4-x\right)& =& 0\\ & & \\ 3x& =& 0 \mathrm{eller} 4-x=0\\ & & \\ x& =& 0 \mathrm{eller} -x=-4\\ & & \\ x_1& =& 0 \mathrm{eller} x_2=4\end{array}$

$\begin{array}{rcl}5x^2-25x& = & 0\\ & & \\ 5x\left(x-5\right)& =& 0\\ & & \\ 5x& =& 0 \mathrm{eller} x-5=0\\ & & \\ x_1& =& 0 \mathrm{eller} x_2=5\end{array}$

$\begin{array}{rcl}{x}^2& = & 4\\ & & \\ x& =& \sqrt{4} \mathrm{eller} x=-\sqrt{4}\\ & & \\ x_1& =& 2 \mathrm{eller} x_2=-2\end{array}$

$\begin{array}{rcl}{x}^2& = & 1\\ & & \\ x& =& \sqrt{1} \mathrm{eller} x=-\sqrt{1}\\ & & \\ x_1& =& 1 \mathrm{eller} x_2=-1\end{array}$

$\begin{array}{rcl}2x^2& = & -8\\ & & \\ x^2& =& -4\\ & & \mathrm{Ingen} \mathrm{løsning}\end{array}$

$\begin{array}{rcl}x-5& = & 4 \mathrm{eller} x-5=-4\\ & & \\ x_1& =& 9 \mathrm{eller} x_2=1\end{array}$

$\begin{array}{rcl}{x}^2+2·3x+3^2& = & 7+9\\ & & \\ x^2+6x+9& =& 16\\ & & \\ {\left(x+3\right)}^2& =& 4^2 \\ & & \\ x+3& =& 4 \mathrm{eller} x+3=-4\\ & & \\ x_1& =& 1 \mathrm{eller} x_2=-7\end{array}$

$\begin{array}{rcl} x^2-2x& = & 24\\ & & \\ x^2-2·x+1^2& =& 24+1\\ & & \\ x^2-2x+1& =& 25\\ & & \\ {\left(x-1\right)}^2& =& 5^2\\ & & \\ x-1& =& 5 \mathrm{eller} x-1=-5\\ & & \\ x_1& =& 6 \mathrm{eller} x_2=-4\end{array}$

$\begin{array}{rcl}-2x^2+4x+16& = & 0 \overline{):\left(-2\right)}\\ & & \\ x^2-2x-8& =& 0\\ & & \\ x^2-2·x+1^2& =& 8+1\\ & & \\ {\left(x-1\right)}^2& =& 3^2\\ & & \\ x-1& =& 3 \mathrm{eller} x-1=-3\\ & & \\ x_1& =& 4 \mathrm{eller} x_2=-2\end{array}$

$\begin{array}{rcl}0,2x^2+0,4x& = & 0,6 \overline{):0,2}\\ & & \\ x^2+2x+1^2& =& 3+1\\ & & \\ {\left(x+1\right)}^2& =& 2^2\\ & & \\ x+1& =& 2 \mathrm{eller} x+1=-2\\ & & \\ x_1& =& 1 \mathrm{eller} x_2 =-3\end{array}$

$\begin{array}{rcl}0,1x^2+0,2x& = & 0,8 \overline{):0,1}\\ & & \\ x^2+2x+1^2& =& 8+1\\ & & \\ {\left(x+1\right)}^2& =& 3^2\\ & & \\ x+1& =& 3 \mathrm{eller} x+1=-3\\ & & \\ x_1& =& 2 \mathrm{eller} x_2=-4\end{array}$