Answer:
[tex]6.39\ miles/hr[/tex]
Step-by-step explanation:
[tex]Let\ her\ speed=x\ mile/hr\\\\Let\ time\ taken=t\ hr\\\\distance=speed\times time\\\\10=x\times t\ .........................(1)[/tex]
When speed is 3 mi/hr faster.
[tex]Her\ speed=x+3\ mile/hr\\\\Time\ taken=t-\frac{1}{2}\ hr\ \ \ as\ it\ takes 30\ minutes\ fewer\ and\ 30\ min=\frac{1}{2}\ hr\\\\distance=speed\times time\\\\10=(x+3)\times (t-\frac{1}{2})\\\\10=xt+3t-\frac{x}{2}-\frac{3}{2}\......................eq(2)\\\\eq(2)-eq(1)\\\\10-10=xt+3t-\frac{x}{2}-\frac{3}{2}-xt\\\\3t=\frac{x}{2}+\frac{3}{2}\\\\t=\frac{x+3}{6}\\\\from\ equation\ 1\ t=\frac{10}{x}\\\\\frac{10}{x}=\frac{x+3}{6}\\\\60=x^2+3x\ \ \ \ \ \ \ \ \ cross\ multiplication\\\\x^2+3x-60=0[/tex]
[tex]Solution\ of\ ax^2+bx+c=0\ is\ given\ by\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\Here\ a=1,b=3\ and c=-60\\\\x=\frac{-3\pm\sqrt{3^2-4\times 1\times (-60)}}{2}\\\\x=\frac{-3\pm\sqrt{9+240}}{2}\\\\Only\ positive\ value\ will\ be\ taken\\\\x=\frac{-3+\sqrt{9+240}}{2}=6.39\ miles/hr[/tex]
