# Question #c2e26

You can first write it into the general form $a {x}^{2} + b x + c = 0$, by noving to the left $x + 7$:
$4 {x}^{2} - x + 4 - x - 7 = 0$ that becomes:
$4 {x}^{2} - 2 x - 3 = 0$
Now it is in the form: $a {x}^{2} + b x + c = 0$ with:
$a = 4$
$b = - 2$
$c = - 3$
${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$