Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-x^{2}+x+20=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx+20$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=-20=-20$$
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 $-20$.
$$-1,20$$ $$-2,10$$ $$-4,5$$
Calculate the sum for each pair.
$$-1+20=19$$ $$-2+10=8$$ $$-4+5=1$$
The solution is the pair that gives sum $1$.
$$a=5$$ $$b=-4$$
Rewrite $-x^{2}+x+20$ as $\left(-x^{2}+5x\right)+\left(-4x+20\right)$.
$$\left(-x^{2}+5x\right)+\left(-4x+20\right)$$
Factor out $-x$ in the first and $-4$ in the second group.
$$-x\left(x-5\right)-4\left(x-5\right)$$
Factor out common term $x-5$ by using distributive property.
$$\left(x-5\right)\left(-x-4\right)$$
To find equation solutions, solve $x-5=0$ and $-x-4=0$.
$$x=5$$ $$x=-4$$
Steps Using the Quadratic Formula
Subtract $x^{2}$ from both sides.
$$5x+20-x^{2}=4x$$
Subtract $4x$ from both sides.
$$5x+20-x^{2}-4x=0$$
Combine $5x$ and $-4x$ to get $x$.
$$x+20-x^{2}=0$$
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.
$$-x^{2}+x+20=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $1$ for $b$, and $20$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{-1±9}{-2}$ when $±$ is plus. Add $-1$ to $9$.
$$x=\frac{8}{-2}$$
Divide $8$ by $-2$.
$$x=-4$$
Now solve the equation $x=\frac{-1±9}{-2}$ when $±$ is minus. Subtract $9$ from $-1$.
$$x=-\frac{10}{-2}$$
Divide $-10$ by $-2$.
$$x=5$$
The equation is now solved.
$$x=-4$$ $$x=5$$
Steps for Completing the Square
Subtract $x^{2}$ from both sides.
$$5x+20-x^{2}=4x$$
Subtract $4x$ from both sides.
$$5x+20-x^{2}-4x=0$$
Combine $5x$ and $-4x$ to get $x$.
$$x+20-x^{2}=0$$
Subtract $20$ from both sides. Anything subtracted from zero gives its negation.
$$x-x^{2}=-20$$
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 $x^{2}+bx=c$.
$$-x^{2}+x=-20$$
Divide both sides by $-1$.
$$\frac{-x^{2}+x}{-1}=-\frac{20}{-1}$$
Dividing by $-1$ undoes the multiplication by $-1$.
$$x^{2}+\frac{1}{-1}x=-\frac{20}{-1}$$
Divide $1$ by $-1$.
$$x^{2}-x=-\frac{20}{-1}$$
Divide $-20$ by $-1$.
$$x^{2}-x=20$$
Divide $-1$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{2}$. Then add the square of $-\frac{1}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
