Solve 4y^2+15y+109geq0

$$4 y ^ { 2 } + 15 y + 109 \geq 0$$

Solve for y

$$y\in \mathrm{R}$$

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.

$$4y^{2}+15y+109=0$$

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 4 for a, 15 for b, and 109 for c in the quadratic formula.

$$y=\frac{-15±\sqrt{15^{2}-4\times 4\times 109}}{2\times 4}$$

Do the calculations.

$$y=\frac{-15±\sqrt{-1519}}{8}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 4y^{2}+15y+109 has the same sign for any y. To determine the sign, calculate the value of the expression for y=0.

$$4\times 0^{2}+15\times 0+109=109$$

The value of the expression 4y^{2}+15y+109 is always positive. Inequality holds for y\in \mathrm{R}.

$$y\in \mathrm{R}$$