Answer
Length $=6.2$ inches.
Width $=1.6$ inches.
Work Step by Step
Let the length be $l$.
and width be $w$.
Area of the rectangle is
$A=l\cdot w$ ... (1)
From the question we have
$l=2w+3$ ... (2)
$A=10$ square inches.
By using equation (1) and (2).
$10=(2w+3)\cdot w$
Simplify.
$10=2w^2+3w$
Subtract $10$ from both sides.
$10-10=2w^2+3w-10$
$0=2w^2+3w-10$
The quadratic equation is
$2w^2+3w-10=0$
By using quadratic equation solution.
$w=\frac{-3 \pm \sqrt {3^2-4(2)(-10)}}{2(2)}$
$w=\frac{-3\pm \sqrt {9+80}}{4}$
$w=\frac{-3\pm \sqrt {89}}{4}$
Take positive value.
$w=\frac{-3+ \sqrt {89}}{4}$
$w=1.60849528301$
Round to the nearest tenth of an inch.
$w=1.6$ inches.
Plug the value of $w$ into equation (2).
$l=2(1.6)+3$
$l=3.2+3$
$l=6.2$ inches.
