Solve 2x^2-6>-3x

$$2 x ^ { 2 } - 6 > - 3 x$$

Solve for x

$$x\in \left(-\infty,\frac{-\sqrt{57}-3}{4}\right)\cup \left(\frac{\sqrt{57}-3}{4},\infty\right)$$

Steps Using the Quadratic Formula

Add 3x to both sides.

$$2x^{2}-6+3x>0$$

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.

$$2x^{2}-6+3x=0$$

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 2 for a, 3 for b, and -6 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{57}}{4}$$

Solve the equation x=\frac{-3±\sqrt{57}}{4} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{57}-3}{4} x=\frac{-\sqrt{57}-3}{4}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{\sqrt{57}-3}{4}\right)\left(x-\frac{-\sqrt{57}-3}{4}\right)>0$$

For the product to be positive, x-\frac{\sqrt{57}-3}{4} and x-\frac{-\sqrt{57}-3}{4} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{57}-3}{4} and x-\frac{-\sqrt{57}-3}{4} are both negative.

$$x-\frac{\sqrt{57}-3}{4}<0 x-\frac{-\sqrt{57}-3}{4}<0$$

The solution satisfying both inequalities is x<\frac{-\sqrt{57}-3}{4}.

$$x<\frac{-\sqrt{57}-3}{4}$$

Consider the case when x-\frac{\sqrt{57}-3}{4} and x-\frac{-\sqrt{57}-3}{4} are both positive.

$$x-\frac{-\sqrt{57}-3}{4}>0 x-\frac{\sqrt{57}-3}{4}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{57}-3}{4}.

$$x>\frac{\sqrt{57}-3}{4}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-\sqrt{57}-3}{4}\text{; }x>\frac{\sqrt{57}-3}{4}$$