Answer
The lengths are $12\; units,9\; units$ and $15\; units$.
Work Step by Step
one leg $a=12$.
other leg $b=x$.
hypotenuse $c=2x-3$.
By using Pythagorean theorem.
$c^2=a^2+b^2$
Plug all values.
$(2x-3)^2=(12)^2+(x)^2$
Clear the parentheses.
$4x^2-12x+9=144+x^2$
Add $-144-x^2$ to both sides.
$4x^2-12x+9-144-x^2=144+x^2-144-x^2$
Simplify.
$3x^2-12x-135=0$
Divide both sides by $3$.
$\frac{3x^2}{3}-\frac{12x}{3}-\frac{135}{3}=0$
Simplify.
$x^2-4x-45=0$
Rewrite the middle term $-4x$ as $-9x+5x$
$x^2-9x+5x-45=0$
Group terms
$(x^2-9x)+(5x-45)=0$
Factor from each term.
$x(x-9)+5(x-9)=0$
Factor out $(x-9)$
$(x-9)(x+5)=0$
Set each factor equal to zero.
$x-9=0$ or $x+5=0$
Isolate $x$.
$x=9$ or $x=-5$
Take the positive value $x=9$.
$c=2x-3$
$c=2(9)-3$
$c=18-3$
$c=15$
