Mathematics

# find the roots ? x² + 7x + 12 =0 x²-8x +12=0

We have to remember one simple pattern $\Delta =b^2-4ac$ 1. $x^2+7x+12=0$ in this case $b=7$ $a=1$ $c=12$ so: $\Delta =7^2-4*1*12$ $\Delta =49-48$ $\Delta =1$ $\Delta >0$ it means that there are 2 roots $\sqrt{\Delta} =1$ $x_{1}=\frac{-b+\sqrt{\Delta}}{2a}$ $x_{2}=\frac{-b-\sqrt{\Delta}}{2a}$ $x_{1}=\frac{-7+1}{2}$ $x_{1}=\frac{-6}{2}$ $x_{1}=-3$ $x_{2}=\frac{-7-1}{2}$ $x_{2}=\frac{-8}{2}$ $x_{2}=-4$ Roots are $x_1=-4$ and $x_2=-3$ 2. $x^2-8x +12=0$ in this case $a=1$ $b=-8$ $c=12$ $\Delta =(-8)^2-4*1*12$ $\Delta =64-48$ $\Delta =16$ $\Delta >0$ it means that there are 2 roots $\sqrt{\Delta}=4$ $x_{1}=\frac{-b+\sqrt{\Delta}}{2a}$ $x_{2}=\frac{-b-\sqrt{\Delta}}{2a}$ $x_{1}=\frac{8+4}{2}$ $x_{1}=\frac{12}{2}$ $x_{1}=6$ $x_{2}=\frac{8-4}{2}$ $x_{2}=\frac{4}{2}$ $x_{2}=2$ Roots:$x_1=6$ and $x_2=4$
$x^2 + 7x + 12 =0\\ x^2+4x+3x+12=0\\ x(x+4)+3(x+4)=0\\ (x+3)(x+4)=0\\ x=-3 \vee x=-4\\\\ x^2-8x+12=0\\ x^2-6x-2x+12=0\\ x(x-6)-2(x-6)=0\\ (x-2)(x-6)=0\\ x=2 \vee x=6$