Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}+4x+3=0$$
To solve the equation, factor $x^{2}+4x+3$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.
$$a+b=4$$ $$ab=3$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.
$$a=1$$ $$b=3$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x+1\right)\left(x+3\right)$$
To find equation solutions, solve $x+1=0$ and $x+3=0$.
$$x=-1$$ $$x=-3$$
Steps Using Factoring By Grouping
Add $x^{2}$ to both sides.
$$4x+3+x^{2}=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}+4x+3=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=4$$ $$ab=1\times 3=3$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.
$$a=1$$ $$b=3$$
Rewrite $x^{2}+4x+3$ as $\left(x^{2}+x\right)+\left(3x+3\right)$.
$$\left(x^{2}+x\right)+\left(3x+3\right)$$
Factor out $x$ in the first and $3$ in the second group.
$$x\left(x+1\right)+3\left(x+1\right)$$
Factor out common term $x+1$ by using distributive property.
$$\left(x+1\right)\left(x+3\right)$$
To find equation solutions, solve $x+1=0$ and $x+3=0$.
$$x=-1$$ $$x=-3$$
Steps Using the Quadratic Formula
Add $x^{2}$ to both sides.
$$4x+3+x^{2}=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$x^{2}+4x+3=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $4$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-4±\sqrt{4^{2}-4\times 3}}{2}$$
Square $4$.
$$x=\frac{-4±\sqrt{16-4\times 3}}{2}$$
Multiply $-4$ times $3$.
$$x=\frac{-4±\sqrt{16-12}}{2}$$
Add $16$ to $-12$.
$$x=\frac{-4±\sqrt{4}}{2}$$
Take the square root of $4$.
$$x=\frac{-4±2}{2}$$
Now solve the equation $x=\frac{-4±2}{2}$ when $±$ is plus. Add $-4$ to $2$.
$$x=-\frac{2}{2}$$
Divide $-2$ by $2$.
$$x=-1$$
Now solve the equation $x=\frac{-4±2}{2}$ when $±$ is minus. Subtract $2$ from $-4$.
$$x=-\frac{6}{2}$$
Divide $-6$ by $2$.
$$x=-3$$
The equation is now solved.
$$x=-1$$ $$x=-3$$
Steps for Completing the Square
Add $x^{2}$ to both sides.
$$4x+3+x^{2}=0$$
Subtract $3$ from both sides. Anything subtracted from zero gives its negation.
$$4x+x^{2}=-3$$
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
$$x^{2}+4x=-3$$
Divide $4$, the coefficient of the $x$ term, by $2$ to get $2$. Then add the square of $2$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}+4x+2^{2}=-3+2^{2}$$
Square $2$.
$$x^{2}+4x+4=-3+4$$
Add $-3$ to $4$.
$$x^{2}+4x+4=1$$
Factor $x^{2}+4x+4$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+2\right)^{2}=1$$
Take the square root of both sides of the equation.
