Question

When the sum of 5 and three times a positive number is subtracted from the square of the number, 0 results. Find the number.

Answer 1

$x-the\ number\\\\x^2-(5+3x)=0\\\\x^2-5-3x=0\\\\x^2-3x-5=0$

$Use\ the\ quadratic\ formula:\\a^2x+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta=0\ then\ x_0=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ then\ no\ solution$

$x^2-3x-5=0\\\\a=1;\ b=-3;\ c=-5\\\\\Delta=(-3)^2-4\cdot1\cdot(-5)+3+20=29 \ \textgreater \ 0\\\\\sqrt\Delta=\sqrt{29}\\\\x_1=\dfrac{3-\sqrt{29}}{2\cdot1}=\dfrac{3-\sqrt{29}}{2} \ \textless \ 0\\\\x_2=\dfrac{3+\sqrt{29}}{2\cdot1}=\dfrac{3+\sqrt{29}}{2} \ \textgreater \ 0$

$Answer:\ \boxed{x=\dfrac{3+\sqrt{29}}{2}}$