Solve x^2+4x-7<0

$$x ^ { 2 } + 4 x - 7 < 0$$

Solve for x

$$x\in \left(-\sqrt{11}-2,\sqrt{11}-2\right)$$

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.

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

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 1 for a, 4 for b, and -7 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\left(-7\right)}}{2}$$

Do the calculations.

$$x=\frac{-4±2\sqrt{11}}{2}$$

Solve the equation x=\frac{-4±2\sqrt{11}}{2} when ± is plus and when ± is minus.

$$x=\sqrt{11}-2 x=-\sqrt{11}-2$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{11}-2\right)\right)\left(x-\left(-\sqrt{11}-2\right)\right)<0$$

For the product to be negative, x-\left(\sqrt{11}-2\right) and x-\left(-\sqrt{11}-2\right) have to be of the opposite signs. Consider the case when x-\left(\sqrt{11}-2\right) is positive and x-\left(-\sqrt{11}-2\right) is negative.

$$x-\left(\sqrt{11}-2\right)>0 x-\left(-\sqrt{11}-2\right)<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(-\sqrt{11}-2\right) is positive and x-\left(\sqrt{11}-2\right) is negative.

$$x-\left(-\sqrt{11}-2\right)>0 x-\left(\sqrt{11}-2\right)<0$$

The solution satisfying both inequalities is x\in \left(-\left(\sqrt{11}+2\right),\sqrt{11}-2\right).

$$x\in \left(-\left(\sqrt{11}+2\right),\sqrt{11}-2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\sqrt{11}-2,\sqrt{11}-2\right)$$