Solve 2(3x+5)>-x^2

$$2 ( 3 x + 5 ) > - x ^ { 2 }$$

Solve for x

$$x\in \mathrm{R}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by 3x+5.

$$6x+10>-x^{2}$$

Add x^{2} to both sides.

$$6x+10+x^{2}>0$$

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.

$$6x+10+x^{2}=0$$

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, 6 for b, and 10 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{-6±\sqrt{-4}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 6x+10+x^{2} has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$6\times 0+10+0^{2}=10$$

The value of the expression 6x+10+x^{2} is always positive. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$