solve the inequality: $-c^{2}leq -c + 2$\ninterval notation solution:\nno solution\nnext question

Explanation:

Step1: Rewrite the inequality

Move all terms to one side: ( -c^{2}+c - 2\leq0), multiply both sides by (- 1) (reverse the inequality sign) to get (c^{2}-c + 2\geq0).

Step2: Find the roots of the corresponding quadratic equation

For the quadratic equation (c^{2}-c + 2 = 0), use the quadratic formula (c=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}), where (a = 1), (b=-1), (c = 2). Then (\Delta=b^{2}-4ac=(-1)^{2}-4\times1\times2=1 - 8=-7<0).

Step3: Analyze the quadratic function

The quadratic function (y = c^{2}-c + 2) has (a = 1>0) (opens upwards) and (\Delta<0). So (y=c^{2}-c + 2\geq0) for all real values of (c).

Answer:

((-\infty,\infty))