-2x^2-x+4=-4-4-4 (4)/(-2)=-6

Let's solve the equation step by step:<br /><br />Given equation: $-2x^{2}-x+4=-4-4-4$<br /><br />First, simplify the right side of the equation:<br />$-2x^{2}-x+4=-12$<br /><br />Next, move all terms to one side of the equation to set it to zero:<br />$-2x^{2}-x+4+12=0$<br />$-2x^{2}-x+16=0$<br /><br />Now, we have a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = -2$, $b = -1$, and $c = 16$.<br /><br />We can solve this quadratic equation using the quadratic formula:<br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />Substitute the values of $a$, $b$, and $c$ into the formula:<br />$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(16)}}{2(-2)}$<br />$x = \frac{1 \pm \sqrt{1 + 128}}{-4}$<br />$x = \frac{1 \pm \sqrt{129}}{-4}$<br /><br />Therefore, the solutions to the equation are:<br />$x = \frac{1 + \sqrt{129}}{-4}$ and $x = \frac{1 - \sqrt{129}}{-4}$<br /><br />Regarding the second expression $\frac{4}{-2}=-6$, it is incorrect. The correct calculation is:<br />$\frac{4}{-2} = -2$<br /><br />So, the correct answer is:<br />$\frac{4}{-2} = -2$