To solve the equation using our calculator with a table, we first have to rewrite the equation in standard form.

10 x^{2} + 3 = 11 x

$LHS - 11 x = RHS - 11 x$

10 x^{2} - 11 x + 3 = 0

Having rewritten the equation in standard form, we can enter it in the calculator by pushing

Y =

and typing it in the first row.

Next, by pushing $2 ND$ and $GRAPH,$ we get a table of values for whole number inputs of $x .$ We are looking for $x -$ values that make the $y -$ column equal $0 .$

From the table, we see that the function decreases at first and starts increasing somewhere between $x = 0$ and $x = 2 .$ If the equation has any solutions, they should be between $x = 0$ and $x = 2 .$ To find these exact values, we will need to view smaller increments in the table. To change the table settings, push $2 ND$ and then $WINDOW .$ Then change the TblStart to $0$ and $△ Tbl$ to $0.1 .$

Now push $2 ND$ and $GRAPH$ to see the table again with the new setup.

Now we can see that $x = 0.5$ and $x = 0.6$ both provide a value of $0$ in the $y -$ column. These are the zeros of the function, so the solutions to the quadratic equation are $x = 0.5$ and $x = 0.6 .$

Exercise