Swap sides so that all variable terms are on the left hand side.
$$-26x-x^{2}=25$$
Subtract $25$ from both sides.
$$-26x-x^{2}-25=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-x^{2}-26x-25=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx-25$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-26$$ $$ab=-\left(-25\right)=25$$
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 $25$.
$$-1,-25$$ $$-5,-5$$
Calculate the sum for each pair.
$$-1-25=-26$$ $$-5-5=-10$$
The solution is the pair that gives sum $-26$.
$$a=-1$$ $$b=-25$$
Rewrite $-x^{2}-26x-25$ as $\left(-x^{2}-x\right)+\left(-25x-25\right)$.
$$\left(-x^{2}-x\right)+\left(-25x-25\right)$$
Factor out $x$ in the first and $25$ in the second group.
$$x\left(-x-1\right)+25\left(-x-1\right)$$
Factor out common term $-x-1$ by using distributive property.
$$\left(-x-1\right)\left(x+25\right)$$
To find equation solutions, solve $-x-1=0$ and $x+25=0$.
$$x=-1$$ $$x=-25$$
Steps Using the Quadratic Formula
Swap sides so that all variable terms are on the left hand side.
$$-26x-x^{2}=25$$
Subtract $25$ from both sides.
$$-26x-x^{2}-25=0$$
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.
$$-x^{2}-26x-25=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-26$ for $b$, and $-25$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{26±24}{-2}$ when $±$ is plus. Add $26$ to $24$.
$$x=\frac{50}{-2}$$
Divide $50$ by $-2$.
$$x=-25$$
Now solve the equation $x=\frac{26±24}{-2}$ when $±$ is minus. Subtract $24$ from $26$.
$$x=\frac{2}{-2}$$
Divide $2$ by $-2$.
$$x=-1$$
The equation is now solved.
$$x=-25$$ $$x=-1$$
Steps for Completing the Square
Swap sides so that all variable terms are on the left hand side.
$$-26x-x^{2}=25$$
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 $x^{2}+bx=c$.
$$-x^{2}-26x=25$$
Divide both sides by $-1$.
$$\frac{-x^{2}-26x}{-1}=\frac{25}{-1}$$
Dividing by $-1$ undoes the multiplication by $-1$.
Divide $26$, the coefficient of the $x$ term, by $2$ to get $13$. Then add the square of $13$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}+26x+13^{2}=-25+13^{2}$$
Square $13$.
$$x^{2}+26x+169=-25+169$$
Add $-25$ to $169$.
$$x^{2}+26x+169=144$$
Factor $x^{2}+26x+169$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+13\right)^{2}=144$$
Take the square root of both sides of the equation.
