Question 1040824
There should be 2 solutions, or roots.
either:
2 imaginary roots
or
2 real roots
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{{{ 4x^2 - 16x + 15 = 0 }}}
I can use the quadratic formula
{{{ x = (-b +- sqrt( b^2-4*a*c )) / (2*a) }}} 
{{{ a = 4 }}}
{{{ b = -16 }}}
{{{ c = 15 }}}
{{{ x = (-(-16) +- sqrt( (-16)^2 - 4*4*15 )) / (2*4) }}} 
{{{ x = ( 16 +- sqrt( 256 - 240 )) / 8 }}} 
{{{ x = ( 16 + sqrt( 16 ) ) / 8 }}}
{{{ x = ( 16 + 4 ) / 8 }}}
{{{ x = 5/2 }}}
and
{{{ x = ( 16 - sqrt( 16 ) ) / 8 }}}
{{{ x = ( 16 - 4 ) / 8 }}}
{{{ x = 3/2 }}}
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check the solutions:
{{{ ( x - 3/2 )*( x - 5/2 ) = 0 }}}
{{{ x^2 -(3/2)*x - (5/2)*x + 15/4 = 0 }}}
{{{ x^2 - 4x + 15/4 = 0 }}}
Multiply both sides by {{{ 4 }}}
{{{ 4x^2 - 16x + 15 = 0 }}}
OK