Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$4s^{2}-12s+9=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $4s^{2}+as+bs+9$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-12$$ $$ab=4\times 9=36$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $36$.
Rewrite $4s^{2}-12s+9$ as $\left(4s^{2}-6s\right)+\left(-6s+9\right)$.
$$\left(4s^{2}-6s\right)+\left(-6s+9\right)$$
Factor out $2s$ in the first and $-3$ in the second group.
$$2s\left(2s-3\right)-3\left(2s-3\right)$$
Factor out common term $2s-3$ by using distributive property.
$$\left(2s-3\right)\left(2s-3\right)$$
Rewrite as a binomial square.
$$\left(2s-3\right)^{2}$$
To find equation solution, solve $2s-3=0$.
$$s=\frac{3}{2}$$
Steps Using the Quadratic Formula
Subtract $12s$ from both sides.
$$4s^{2}+9-12s=0$$
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.
$$4s^{2}-12s+9=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $4$ for $a$, $-12$ for $b$, and $9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Divide $-3$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{2}$. Then add the square of $-\frac{3}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
