Given the figure of a rectangle
The area = A = 26
Length = x + 6
width = x + 2
Area = length * Width
so, the equation of the area will be:
[tex]A=(x+6)(x+2)[/tex]
so,
[tex](x+6)(x+2)=26[/tex]
solve for x as follows:
[tex]\begin{gathered} x^2+8x+12=26 \\ x^2+8x+12-26=0 \\ x^2+8x-14=0 \\ \end{gathered}[/tex]
Use the general rule to find the value of x
So,
[tex]\begin{gathered} a=1,b=8,c=-14 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}=\frac{-8\pm\sqrt[]{64-4\cdot1\cdot-14}}{2\cdot1} \\ \\ x=\frac{-8\pm\sqrt[]{120}}{2}=\frac{-8\pm2\sqrt[]{30}}{2}=-4\pm\sqrt[]{30} \end{gathered}[/tex]
So, the answer will be:
[tex]\begin{gathered} A=(x+6)(x+2)_{} \\ \\ x=-4+\sqrt[]{30},-4-\sqrt[]{30} \end{gathered}[/tex]
