Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.
$$xx-4=-3x$$
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}-4=-3x$$
Add $3x$ to both sides.
$$x^{2}-4+3x=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}+3x-4=0$$
To solve the equation, factor $x^{2}+3x-4$ 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.
$$a+b=3$$ $$ab=-4$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-4$.
$$-1,4$$ $$-2,2$$
Calculate the sum for each pair.
$$-1+4=3$$ $$-2+2=0$$
The solution is the pair that gives sum $3$.
$$a=-1$$ $$b=4$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-1\right)\left(x+4\right)$$
To find equation solutions, solve $x-1=0$ and $x+4=0$.
$$x=1$$ $$x=-4$$
Steps Using Factoring By Grouping
Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.
$$xx-4=-3x$$
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}-4=-3x$$
Add $3x$ to both sides.
$$x^{2}-4+3x=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$x^{2}+3x-4=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-4$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=1\left(-4\right)=-4$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-4$.
$$-1,4$$ $$-2,2$$
Calculate the sum for each pair.
$$-1+4=3$$ $$-2+2=0$$
The solution is the pair that gives sum $3$.
$$a=-1$$ $$b=4$$
Rewrite $x^{2}+3x-4$ as $\left(x^{2}-x\right)+\left(4x-4\right)$.
$$\left(x^{2}-x\right)+\left(4x-4\right)$$
Factor out $x$ in the first and $4$ in the second group.
$$x\left(x-1\right)+4\left(x-1\right)$$
Factor out common term $x-1$ by using distributive property.
$$\left(x-1\right)\left(x+4\right)$$
To find equation solutions, solve $x-1=0$ and $x+4=0$.
$$x=1$$ $$x=-4$$
Steps Using the Quadratic Formula
Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.
$$xx-4=-3x$$
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}-4=-3x$$
Add $3x$ to both sides.
$$x^{2}-4+3x=0$$
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.
$$x^{2}+3x-4=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $3$ for $b$, and $-4$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-3±\sqrt{3^{2}-4\left(-4\right)}}{2}$$
Square $3$.
$$x=\frac{-3±\sqrt{9-4\left(-4\right)}}{2}$$
Multiply $-4$ times $-4$.
$$x=\frac{-3±\sqrt{9+16}}{2}$$
Add $9$ to $16$.
$$x=\frac{-3±\sqrt{25}}{2}$$
Take the square root of $25$.
$$x=\frac{-3±5}{2}$$
Now solve the equation $x=\frac{-3±5}{2}$ when $±$ is plus. Add $-3$ to $5$.
$$x=\frac{2}{2}$$
Divide $2$ by $2$.
$$x=1$$
Now solve the equation $x=\frac{-3±5}{2}$ when $±$ is minus. Subtract $5$ from $-3$.
$$x=-\frac{8}{2}$$
Divide $-8$ by $2$.
$$x=-4$$
The equation is now solved.
$$x=1$$ $$x=-4$$
Steps for Completing the Square
Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.
$$xx-4=-3x$$
Multiply $x$ and $x$ to get $x^{2}$.
$$x^{2}-4=-3x$$
Add $3x$ to both sides.
$$x^{2}-4+3x=0$$
Add $4$ to both sides. Anything plus zero gives itself.
$$x^{2}+3x=4$$
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.
