Solve 23x^2+12x-35geq0

$$23 x ^ { 2 } + 12 x - 35 \geq 0$$

Solve for x

$$x\in (-\infty,-\frac{35}{23}]\cup [1,\infty)$$

Steps Using the Quadratic Formula

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.

$$23x^{2}+12x-35=0$$

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 23 for a, 12 for b, and -35 for c in the quadratic formula.

$$x=\frac{-12±\sqrt{12^{2}-4\times 23\left(-35\right)}}{2\times 23}$$

Do the calculations.

$$x=\frac{-12±58}{46}$$

Solve the equation x=\frac{-12±58}{46} when ± is plus and when ± is minus.

$$x=1 x=-\frac{35}{23}$$

Rewrite the inequality by using the obtained solutions.

$$23\left(x-1\right)\left(x+\frac{35}{23}\right)\geq 0$$

For the product to be ≥0, x-1 and x+\frac{35}{23} have to be both ≤0 or both ≥0. Consider the case when x-1 and x+\frac{35}{23} are both ≤0.

$$x-1\leq 0 x+\frac{35}{23}\leq 0$$

The solution satisfying both inequalities is x\leq -\frac{35}{23}.

$$x\leq -\frac{35}{23}$$

Consider the case when x-1 and x+\frac{35}{23} are both ≥0.

$$x+\frac{35}{23}\geq 0 x-1\geq 0$$

The solution satisfying both inequalities is x\geq 1.

$$x\geq 1$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{35}{23}\text{; }x\geq 1$$