Solve 4x^2+6x-7leq0

$$4 x ^ { 2 } + 6 x - 7 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}\frac{-\sqrt{37}-3}{4},\frac{\sqrt{37}-3}{4}\end{bmatrix}$$

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.

$$4x^{2}+6x-7=0$$

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, 6 for b, and -7 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-7\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{-6±2\sqrt{37}}{8}$$

Solve the equation x=\frac{-6±2\sqrt{37}}{8} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{37}-3}{4} x=\frac{-\sqrt{37}-3}{4}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{\sqrt{37}-3}{4}\right)\left(x-\frac{-\sqrt{37}-3}{4}\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{\sqrt{37}-3}{4} and x-\frac{-\sqrt{37}-3}{4} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{\sqrt{37}-3}{4}\geq 0 and x-\frac{-\sqrt{37}-3}{4}\leq 0.

$$x-\frac{\sqrt{37}-3}{4}\geq 0 x-\frac{-\sqrt{37}-3}{4}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{\sqrt{37}-3}{4}\leq 0 and x-\frac{-\sqrt{37}-3}{4}\geq 0.

$$x-\frac{-\sqrt{37}-3}{4}\geq 0 x-\frac{\sqrt{37}-3}{4}\leq 0$$

The solution satisfying both inequalities is x\in \left[\frac{-\sqrt{37}-3}{4},\frac{\sqrt{37}-3}{4}\right].

$$x\in \begin{bmatrix}\frac{-\sqrt{37}-3}{4},\frac{\sqrt{37}-3}{4}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}\frac{-\sqrt{37}-3}{4},\frac{\sqrt{37}-3}{4}\end{bmatrix}$$