Question 696339
{{{ 4x^2+ 7x = 15  }}}
{{{ x^2 + (7/4)*x = 15/4 }}}
This is now in the right form to complete the square
Take 1/2 of the co-efficient of the {{{ x }}} term,
square it, then add it to both sides.
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{{{ x^2 + (7/4)*x + ( 7/8)^2 = 15/4 + (7/8)^2 }}}
{{{ x^2 + (7/4)*x + 49/64 = 240/64 + 49/64 }}}
{{{ x^2 + (7/4)*x + 49/64 = 289/64 }}}
{{{ ( x + 7/8 )^2 = (17/8)^2 }}}
Take the square root of both sides
{{{ x + 7/8 = 17/8 }}}
{{{ x = 10/8 }}}
{{{ x = 5/4 }}}
And also take the negative square root of {{{ (17/8)^2 }}}
{{{ x + 7/8 = -17/8 }}}
{{{ x = -24/8 }}}
{{{ x = -3 }}}
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The roots are {{{ 5/4 }}} and {{{ -3 }}}
The factoring is
{{{ ( x + 3 )*( x - 5/4 ) = 4x^2 + 7x - 15 }}}
check:
{{{ ( x + 3 )*( x - 5/4 ) = x^2 + 3x - (5/4)*x - 15/4 }}}
{{{ x^2 + 3x - (5/4)*x - 15/4  = 0 }}}
Multiply both sides by {{{ 4 }}}
{{{ 4x^2 + 12x - 5x - 15 = 0 }}}
{{{ 4x^2 + 7x - 15 = 0 }}}
OK
Here's the plot
{{{ graph( 400, 400, -5, 5, -20, 5, 4x^2 + 7x - 15 ) }}}