Using the quadratic formula solver Example 4

Solve $2x^2-7x+6=0$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Therefore $a=2$ , $b=-7$ & $c=6$

$x=(-(-7)+-sqrt((-7)^2-4times2times6))/(2times2)$

$x=(7+-sqrt(49-48))/4$

$x=(7+-sqrt1)/4$

$x=(7+-1)/4$

$x=(7+1)/4$ or $x=(7-1)/4$

$x=8/4$ or $x=6/4$

$x=2$ or $x=1\1/2$

$2x^2-7x+6=0$

If $x=2$ $2times2^2-7times2+6=0$

$8-14+6=0$ Which is correct

If $x=1\1/2$ $2times(3/2)^2-7times(3/2)+6=0$

$2times(9/4)-21/2+6=0$

$18/4-21/2+6=0$

$4\2/4-10\1/2+6=0$

$4\1/2-10\1/2+6=0$ Which is correct

Answer:

The roots of $2x^2-7x+6=0$ are $x=2$ or $x=1\1/2$

Quick stetch example 4