Solve 4+3x-x^2>0

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

Solve for x

$$x\in \left(-1,4\right)$$

Steps Using the Quadratic Formula

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

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

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.

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

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

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

Do the calculations.

$$x=\frac{3±5}{2}$$

Solve the equation x=\frac{3±5}{2} when ± is plus and when ± is minus.

$$x=4 x=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-4\right)\left(x+1\right)<0$$

For the product to be negative, x-4 and x+1 have to be of the opposite signs. Consider the case when x-4 is positive and x+1 is negative.

$$x-4>0 x+1<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+1 is positive and x-4 is negative.

$$x+1>0 x-4<0$$

The solution satisfying both inequalities is x\in \left(-1,4\right).

$$x\in \left(-1,4\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-1,4\right)$$