Answer:
[tex]Solution,\\Let~h=3x-4~,p=x~and~b=2x+4\\Then,\\h^{2} = p^{2}+b^{2}\\or, (3x-4)^{2}=x^{2}+(2x+4)^{2}\\or, 9x^{2} - 2(3x)(4) +16 = x^{2} +4x^{2}+2(4x)(4)+16\\or, 9x^{2} -24x+16 = 5x^{2}+16x+16\\or, 9x^{2}-5x^{2}-24x-16x=0\\or, 4x^2-40x = 0\\or, 4x(x-10) = 0\\i.e. x = 0~or~x=10\\[/tex]