Explanation
[tex]A=4x^2+8x+9[/tex]Step 1
Let
A=96 (square inches)
so,replacing
[tex]\begin{gathered} A=4x^2+8x+9 \\ 69=4x^2+8x+9 \\ \text{now, to set the equation equal to zero subtract 69 in both sides} \\ 69-69=4x^2+8x+9-69 \\ 0=4x^2+8x-60\Rightarrow equation\text{ (1)} \end{gathered}[/tex]Step 2
now, we need to solve the quadratic equation,we can use the quadratic formula
[tex]\begin{gathered} 4x^2+8x-60=0\Rightarrow ax^2+bx+c \\ thus \\ a=4 \\ b=8 \\ c=-60 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{replace} \\ x=\frac{-8\pm\sqrt[]{8^2-4(4)(-60)}}{2(4)} \\ x=\frac{-8\pm\sqrt[]{1021}}{8} \\ x=\frac{-8\pm32}{8} \end{gathered}[/tex]therefore, the solutions are
[tex]\begin{gathered} x=\frac{-8\pm32}{8} \\ x_1=\frac{-8+32}{8}=\frac{24}{8}=3 \\ x_2=\frac{-8-32}{8}=\frac{-40}{8}=-5 \end{gathered}[/tex]as we are looking for a distance, the ONLY valid answer must be positive
so, the answer is
x=3
I hope this helps you
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