Solve (y^2+25y+4)

$$( y ^ { 2 } + 25 y + 4 )$$

Factor

$$\left(y-\frac{-\sqrt{609}-25}{2}\right)\left(y-\frac{\sqrt{609}-25}{2}\right)$$

Steps Using the Quadratic Formula

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.

$$y^{2}+25y+4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y=\frac{-25±\sqrt{25^{2}-4\times 4}}{2}$$

Square 25.

$$y=\frac{-25±\sqrt{625-4\times 4}}{2}$$

Multiply -4 times 4.

$$y=\frac{-25±\sqrt{625-16}}{2}$$

Add 625 to -16.

$$y=\frac{-25±\sqrt{609}}{2}$$

Now solve the equation y=\frac{-25±\sqrt{609}}{2} when ± is plus. Add -25 to \sqrt{609}.

$$y=\frac{\sqrt{609}-25}{2}$$

Now solve the equation y=\frac{-25±\sqrt{609}}{2} when ± is minus. Subtract \sqrt{609} from -25.

$$y=\frac{-\sqrt{609}-25}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-25+\sqrt{609}}{2} for x_{1} and \frac{-25-\sqrt{609}}{2} for x_{2}.

$$y^{2}+25y+4=\left(y-\frac{\sqrt{609}-25}{2}\right)\left(y-\frac{-\sqrt{609}-25}{2}\right)$$

Evaluate

$$y^{2}+25y+4$$