Solve 6-x^2+4x-3geq0

$$6 - x ^ { 2 } + 4 x - 3 \geq 0$$

Solve for x

$$x\in \begin{bmatrix}2-\sqrt{7},\sqrt{7}+2\end{bmatrix}$$

Steps Using the Quadratic Formula

Subtract 3 from 6 to get 3.

$$3-x^{2}+4x\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 3-x^{2}+4x positive. Since -1 is negative, the inequality direction is changed.

$$-3+x^{2}-4x\leq 0$$

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.

$$-3+x^{2}-4x=0$$

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, -4 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{4±2\sqrt{7}}{2}$$

Solve the equation x=\frac{4±2\sqrt{7}}{2} when ± is plus and when ± is minus.

$$x=\sqrt{7}+2 x=2-\sqrt{7}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{7}+2\right)\right)\left(x-\left(2-\sqrt{7}\right)\right)\leq 0$$

For the product to be ≤0, one of the values x-\left(\sqrt{7}+2\right) and x-\left(2-\sqrt{7}\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(\sqrt{7}+2\right)\geq 0 and x-\left(2-\sqrt{7}\right)\leq 0.

$$x-\left(\sqrt{7}+2\right)\geq 0 x-\left(2-\sqrt{7}\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(\sqrt{7}+2\right)\leq 0 and x-\left(2-\sqrt{7}\right)\geq 0.

$$x-\left(2-\sqrt{7}\right)\geq 0 x-\left(\sqrt{7}+2\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[2-\sqrt{7},\sqrt{7}+2\right].

$$x\in \begin{bmatrix}2-\sqrt{7},\sqrt{7}+2\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}2-\sqrt{7},\sqrt{7}+2\end{bmatrix}$$