Solve left(-2a+10right)^2-4left(2a^2-2a-2right)geq0

$${ \left(-2a+10 \right) }^{ 2 } -4 \left( 2 { a }^{ 2 } -2a-2 \right) \geq 0$$

Solve for a

$$a\in \begin{bmatrix}-\sqrt{43}-4,\sqrt{43}-4\end{bmatrix}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2a+10\right)^{2}.

$$4a^{2}-40a+100-4\left(2a^{2}-2a-2\right)\geq 0$$

Use the distributive property to multiply -4 by 2a^{2}-2a-2.

$$4a^{2}-40a+100-8a^{2}+8a+8\geq 0$$

Combine 4a^{2} and -8a^{2} to get -4a^{2}.

$$-4a^{2}-40a+100+8a+8\geq 0$$

Combine -40a and 8a to get -32a.

$$-4a^{2}-32a+100+8\geq 0$$

Add 100 and 8 to get 108.

$$-4a^{2}-32a+108\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -4a^{2}-32a+108 positive. Since -1 is negative, the inequality direction is changed.

$$4a^{2}+32a-108\leq 0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$4a^{2}+32a-108=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, 32 for b, and -108 for c in the quadratic formula.

$$a=\frac{-32±\sqrt{32^{2}-4\times 4\left(-108\right)}}{2\times 4}$$

Do the calculations.

$$a=\frac{-32±8\sqrt{43}}{8}$$

Solve the equation a=\frac{-32±8\sqrt{43}}{8} when ± is plus and when ± is minus.

$$a=\sqrt{43}-4 a=-\sqrt{43}-4$$

Rewrite the inequality by using the obtained solutions.

$$4\left(a-\left(\sqrt{43}-4\right)\right)\left(a-\left(-\sqrt{43}-4\right)\right)\leq 0$$

For the product to be ≤0, one of the values a-\left(\sqrt{43}-4\right) and a-\left(-\sqrt{43}-4\right) has to be ≥0 and the other has to be ≤0. Consider the case when a-\left(\sqrt{43}-4\right)\geq 0 and a-\left(-\sqrt{43}-4\right)\leq 0.

$$a-\left(\sqrt{43}-4\right)\geq 0 a-\left(-\sqrt{43}-4\right)\leq 0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a-\left(\sqrt{43}-4\right)\leq 0 and a-\left(-\sqrt{43}-4\right)\geq 0.

$$a-\left(-\sqrt{43}-4\right)\geq 0 a-\left(\sqrt{43}-4\right)\leq 0$$

The solution satisfying both inequalities is a\in \left[-\left(\sqrt{43}+4\right),\sqrt{43}-4\right].

$$a\in \begin{bmatrix}-\left(\sqrt{43}+4\right),\sqrt{43}-4\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$a\in \begin{bmatrix}-\sqrt{43}-4,\sqrt{43}-4\end{bmatrix}$$