$$\textit{Solve : }x^2-11x+28=0$$
This is a quarratic equation. This is a quadratic equation because the left hand side of the equatin is a polynocial of degree 2. Lets try to solve this equation in different ways.
Normally you are advsied to do factorization of the expression in the form of (x-a)(x-b)=0.
This will lead to 2 possible solutions namely x-a=0 i.e. x=a or x-b=0 i.e. y=0.
$$\boldsymbol A)\;Solution \;by \; Factorization:\;$$
$$x^2-11x+28\;can\;be\;factorised\;as\;(x-7)(x-4).\\Hence\;the\;equation\;be\;written\;as\;(x-7)(x-4)=0\;\\either\;x-7\;=\;0\;or\;x-4\;=\;0\\i.e.\;either\;x=4\;or\;x=7\;$$
$$\boldsymbol B)\;Solution \;by \; Sridhar\; Acharya\; Method:\;$$
This method of solving any quadratic equation is called Sridhar Acharya method after the name of the famous bengali mathematician.
Sridhar Acharya method provides roots for any quadratic equation of the form:
$$ax^2+bx+c=0\;where\;a.b,c\;are\;real\;and\;a\neq0\textit{and the roots are given by}$$
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
So, if you compare the given equation with Sridhar Acharya equation, a=1. b=-11 and c=28.
$$So,\;the\;roots\;are\;given\;by\;x=\;\frac{-(-11)\pm\sqrt{{(-11)}^2-4.1.(28)}}{2.1}$$
$$i.e.\;x=\;\frac{11\pm\sqrt{121-112}}{2.1}$$
$$i.e.\;x=\;\frac{11\pm\sqrt{9}}2$$
$$i.e.\;x=\;\frac{11\pm3}2$$
$$i.e.\;the\;2\;solutions\;are\;given\;by\;x=\;\frac{11+3}2\;and\;x=\;\frac{11-3}2$$
$$i.e.\;x=\;7\;and\;x=4$$
$$Note\;on\;we\;got\;same\;solution.\;$$
$$Note\;on\;Sridhar\;Acharya\;Method:$$
$$This\;is\;called\;discriminent\;because\;the\;nature\;of\;the\;roots\;of\;the\;quadratic\\equation\;depends\;on\;the\;value\;of\;D.$$
$$There\;can\;be\;3\;situations:\\1.\;D=0,\;the\;roots\;are\;real\;and\;equal\\2.\;D>0,\;roots\;are\;real\;and\;unequal\\3.\;D<0,\;the\;roots\;are\;imaginary\;and\;unequal\\Since \;D=217\;the\; equation \;has\;2\;real\;unequal\;roots.\;$$
$$\boldsymbol C)\;Solution \;by \; Factor\; Theorem:\;$$
$$Factor\;theorem\;states\;that\;if\;f(a)=0,\;then\;x-a\;is\;a\;factor\;of\;f(x).\\Let\;f(x)=\;x^2-11x+28.\\Note\;f(4)=4^2-11.4+28\;=\;0\\Hence\;x-4\;is\;a\;factor\;of\;f(x).\\Similarly\;x-7\;is\;a\;factor.\\So,\;f(x)=\;x^2-11x+28=(x-4)(x-8).\\so,\;f(x)=\;0\Rightarrow(x-4)(x-7)=0\\either\;x=4\;or\;x=7\;$$