Question 1091043
The standard form of a quadratic function is {{{ax^2+bx+c=0}}} where  
{{{ax^2}}} is the quadratic term 

{{{bx}}} is the linear term

{{{c}}} is the constant term

discriminant {{{b^2-4ac}}} tells us how many and what kind of roots quadratic function have 
if {{{b^2-4ac>0}}} quadratic function have two real roots
if {{{b^2-4ac=0}}} quadratic function have one double root
if {{{b^2-4ac<0}}} quadratic function have two complex roots


let check discriminant in your case 

{{{x^2-12x+144=0}}}...... {{{b^2-4ac}}}->{{{(-12)^2-4*1*144}}}->{{{144-576}}}->{{{-432<0}}}; so, your quadratic function will have have two complex roots

use quadratic formula to find roots: 

1.

{{{x^2-12x+144=0}}}

{{{x = (-(-12) +- sqrt( (-12)^2-4*1*144 ))/(2*1) }}}

{{{x = (12 +- sqrt( -432 ))/2 }}}

{{{x = (12 +- i*sqrt( 432 ))/2 }}}

{{{x = (12 +- i*12 sqrt(3))/2 }}}

{{{x = (cross(12)6 +- i*cross(12)6 sqrt(3))/cross(2) }}}

{{{x = 6 + 6i*sqrt(3)}}}

{{{x = 6 - 6i*sqrt(3)}}}


{{{ graph( 600, 600, -10, 200, -10, 200, x^2-12x+144) }}}



2.

{{{x^2+3x-28=0}}}
let check discriminant in this case:

{{{b^2-4ac}}}->{{{3^2-4*1*(-28)}}}->{{{9+112}}}->{{{121>0}}}; so, your quadratic function will have have two real roots


{{{x^2+3x-28=0}}}....factor completely

{{{x^2+7x-4x-28=0}}}

{{{(x^2+7x)-(4x+28)=0}}}

{{{x(x+7)-4(x+7)=0}}}

{{{(x - 4) (x + 7) = 0}}}

solutions:

if {{{(x - 4)  = 0}}}-> {{{x=4}}}
if {{{ (x + 7) = 0}}}-> {{{x=-7}}}

{{{ graph( 600, 600, -30, 30, -30, 30, x^2+3x-28) }}}