Please help me!

$4x^2 - 31x = 8$

                                      Subtract  8  from both sides of the equation.

$4x^2 - 31x - 8 = 0$

Now we can solve this quadratic equation using the quadratic formula.

The quadratic formula says...

If     $ax^2+bx+c=0$    then    $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

So...

If    $4x^2 - 31x - 8 = 0$    then    $x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)}$

$x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)} \\~\\ x = {31 \pm \sqrt{961+128} \over 8} \\~\\ x = {31 \pm \sqrt{1089} \over 8} \\~\\ x = {31 \pm 33 \over 8} \\~\\ \begin{array} \ x=\frac{31+33}{8}\qquad&\text{or}&\qquad x=\frac{31-33}{8} \\~\\ x=\frac{64}{8}&\text{or}&\qquad x=\frac{-2}{8} \\~\\ x=8&\text{or}&\qquad x=-\frac{1}{4} \end{array}$