# Factorize the quadratic equation. y^{2} + 9 = -9y

## Question:

{eq}y^{2} + 9 = -9y {/eq}

The quadratic equation is mathematically expressed in the standard form of ax^2+bx+c=0. The quadratic equation can be used to evaluate the factors or the complex root of the quadratic equation by using the factorization or the quadratic formula.

Given data:

• The given quadratic equation is {eq}\begin{align*} {y^2} + 9 &= - 9y\\ {y^2} + 9y + 9 &= 0 \end{align*} {/eq}

The expression for finding the factors of the quadratic equation is

{eq}{y_{1,2}} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} {/eq}

• Here {eq}a = 1,\,b = 9,\,c = 9 {/eq} are the coefficients of the quadratic equation {eq}{y^2} + 9y + 9 = 0 {/eq} .

Substituting the values in the above expression as,

{eq}\begin{align*} {y_{1,2}} &= \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ {y_{1,2}} &= \dfrac{{ - 9 \pm \sqrt {{{\left( 9 \right)}^2} - 4 \times 1 \times 9} }}{{2 \times 1}}\\ {y_{1,2}} &= \dfrac{{ - 9 \pm 3\sqrt 5 }}{2}\\ y &= \dfrac{{ - 9 + 3\sqrt 5 }}{2},\dfrac{{ - 9 - 3\sqrt 5 }}{2} \end{align*} {/eq}

Thus the factors of the quadratic equation {eq}{y^2} + 9y + 9 = 0 {/eq} is {eq}y = \dfrac{{ - 9 + 3\sqrt 5 }}{2},\dfrac{{ - 9 - 3\sqrt 5 }}{2} {/eq}