$$x^{2}+x-56=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $1$ for $b$, and $-56$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.$$x=\frac{-1±\sqrt{1^{2}-4\left(-56\right)}}{2}$$
Square $1$.$$x=\frac{-1±\sqrt{1-4\left(-56\right)}}{2}$$
Multiply $-4$ times $-56$.$$x=\frac{-1±\sqrt{1+224}}{2}$$
Add $1$ to $224$.$$x=\frac{-1±\sqrt{225}}{2}$$
Take the square root of $225$.$$x=\frac{-1±15}{2}$$
Now solve the equation $x=\frac{-1±15}{2}$ when $±$ is plus. Add $-1$ to $15$.$$x=\frac{14}{2}$$
Divide $14$ by $2$.$$x=7$$
Now solve the equation $x=\frac{-1±15}{2}$ when $±$ is minus. Subtract $15$ from $-1$.$$x=\frac{-16}{2}$$
Divide $-16$ by $2$.$$x=-8$$
The equation is now solved.$$x=7$$ $$x=-8$$