Solve left(x+5right)^2+4(3x-4)geq0

$${ \left(x+5 \right) }^{ 2 } +4(3x-4) \geq 0$$

Solve for x

$$x\in (-\infty,-4\sqrt{7}-11]\cup [4\sqrt{7}-11,\infty)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25+4\left(3x-4\right)\geq 0$$

Use the distributive property to multiply 4 by 3x-4.

$$x^{2}+10x+25+12x-16\geq 0$$

Combine 10x and 12x to get 22x.

$$x^{2}+22x+25-16\geq 0$$

Subtract 16 from 25 to get 9.

$$x^{2}+22x+9\geq 0$$

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.

$$x^{2}+22x+9=0$$

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 1 for a, 22 for b, and 9 for c in the quadratic formula.

$$x=\frac{-22±\sqrt{22^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-22±8\sqrt{7}}{2}$$

Solve the equation x=\frac{-22±8\sqrt{7}}{2} when ± is plus and when ± is minus.

$$x=4\sqrt{7}-11 x=-4\sqrt{7}-11$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(4\sqrt{7}-11\right)\right)\left(x-\left(-4\sqrt{7}-11\right)\right)\geq 0$$

For the product to be ≥0, x-\left(4\sqrt{7}-11\right) and x-\left(-4\sqrt{7}-11\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(4\sqrt{7}-11\right) and x-\left(-4\sqrt{7}-11\right) are both ≤0.

$$x-\left(4\sqrt{7}-11\right)\leq 0 x-\left(-4\sqrt{7}-11\right)\leq 0$$

The solution satisfying both inequalities is x\leq -4\sqrt{7}-11.

$$x\leq -4\sqrt{7}-11$$

Consider the case when x-\left(4\sqrt{7}-11\right) and x-\left(-4\sqrt{7}-11\right) are both ≥0.

$$x-\left(-4\sqrt{7}-11\right)\geq 0 x-\left(4\sqrt{7}-11\right)\geq 0$$

The solution satisfying both inequalities is x\geq 4\sqrt{7}-11.

$$x\geq 4\sqrt{7}-11$$

The final solution is the union of the obtained solutions.

$$x\leq -4\sqrt{7}-11\text{; }x\geq 4\sqrt{7}-11$$