Question 107041
{{{2x^2-5x=3}}}
{{{2x^2-5x-3=3-3}}}
{{{2x^2-5x-3=0}}}

The quadratic formula for solving the value of x in {{{ax^2+bx+c=0}}} is

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


Here, a=2, b=-5, and c=-3.So,

{{{x=(-(-5)+-sqrt((-5)^2-4*2*(-3)))/(2*2)}}}
{{{x=(5+-sqrt(25-(-24)))/(4)}}}
{{{x=(5+-sqrt(25+24))/(4)}}}
{{{x=(5+-sqrt(49))/(4)}}}
{{{x=(5+- 7)/(4)}}}

{{{x=(5+7)/(4)}}} or {{{x=(5-7)/(4)}}}
{{{x=12/4}}} or {{{x=-2/4}}}
{{{x=3}}} or {{{x=-1/2}}}


For your next problem, {{{3x^2 -2x+1 = 0}}}


Here, a=3, b=-2, c=1.


Using the quadratic formula,

{{{x=(-(-2)+-sqrt((-2)^2-4*3*1))/(2*3)}}}
{{{x=(2+-sqrt(4-12))/(6)}}}
{{{x=(2+-sqrt(-8))/(6)}}}

We know that square root of a negative number does not exist because all squares are positive numbers. In this cases, oftentimes, mathematicians use the symbol {{{i}}} for the square root of {{{-1}}}

Let's continue


{{{x=(2+-sqrt((-1)(4)(2)))/(6)}}}
{{{x=(2+- (sqrt(-1))(sqrt(4))(sqrt(2)))/(6)}}}
{{{x=(2+- (i)(2)(sqrt(2)))/(6)}}}
{{{x=(2+- 2i (sqrt(2)))/(6)}}}

Roots involving {{{i}}} are called imaginary roots.


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