Question 471220


Solve:..to do it, simplest way is to use quadratic formula

1.

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

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...note that {{{a=1}}},{{{b=2}}} and {{{c=-4}}}


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


{{{x = (-2 +- sqrt( 4+16 ))/2 }}}


{{{x = (-2 +- sqrt(20 ))/2 }}}


{{{x = (-2 +- 4.47)/2 }}}

solutions:

{{{x = (-2 +4.47)/2 }}}


{{{x = (2.47)/2 }}}

{{{x = 1.235 }}}

or


{{{x = (-2 -4.47)/2 }}}


{{{x = (-6.47)/2 }}}

{{{x = -3.235 }}}

2.


{{{b^2+3b-10=0}}}

{{{b = (-3 +- sqrt( 3^2-4*1*(-10) ))/(2*1) }}} 


{{{b = (-3 +- sqrt( 9+40 ))/2 }}}

{{{b = (-3 +- sqrt(49 ))/2 }}}

{{{b = (-3 +- 7)/2 }}}

solutions:

{{{b = (-3 + 7)/2 }}}

{{{b = 4/2 }}}

{{{b = 2 }}}

or

{{{b = (-3 - 7)/2 }}}

{{{b =-10/2 }}}

{{{b = -5 }}}

3.


{{{r^2-5r-24=0}}}


{{{r = (-(-5) +- sqrt( (-5)^2-4*1*(-24) ))/(2*1) }}}


{{{r = (5 +- sqrt( 25+96 ))/2 }}}


{{{r = (5 +- sqrt( 121))/2 }}}


{{{r = (5 +- 11)/2 }}}


solutions:


{{{r = (5 +11)/2 }}}


{{{r = 16/2 }}}


{{{r = 8 }}}

or


{{{r = (5 -11)/2 }}}


{{{r = -6/2 }}}


{{{r = -3 }}}