Question 86665
Solution:  Always write in form of "(ax2 + bx + c)=0".
So you can use
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} or {{{(x+a)(x+b)=0 }}}
to solve

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#1 {{{2x^2-7x=5 }}} 
{{{2x^2-7x-5=0 }}}
{{{x = (-(-7) +- sqrt( (-7)^2-4*2*(-5) ))/(2*2) }}}
{{{x = ( 7 +- sqrt( 49+40 ))/(4) }}}  
{{{x = ( 7 +- sqrt( 89 ))/(4) }}}
{{{x = ( 7 +- 9.43))/(4) }}}
{{{x = ( 7 + 9.43))/(4) }}} = {{{4.1075 }}}
and
{{{x = (7 - 9.43))/(4) }}} = {{{-0.6075}}}
****Are you sure you copied the equation right?***

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 #2  {{{x^2=x+4 }}}
{{{x^2-x-4=0 }}}
{{{x^2-x-4=0 }}}
{{{x = (-(-1) +- sqrt( (-1)^2-4*1*(-4) ))/(2*1) }}}
{{{x = (1 +- sqrt( 1+16 ))/(2) }}}
{{{x = (1 +- sqrt( 17 ))/(2) }}}
{{{x = (1 +- 4.123)/(2) }}}
{{{x = 2.5615 }}} and {{{x = -1.5615 }}}

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#3)  Find the distance between (7,0) and (-7,0)
Draw right-triangle 

{{{ graph( 300, 300, -10, 10, -10, 10, x+7) }}}

Distance is {{{c^2 = a^2+b^2 }}}

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#4. Solve...  {{{x^2+8=17 }}}
same as #1, and #2 use {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}