Solve for: x ^ 2+x-56 = 0

Expression: $$x ^ { 2 } + x - 56 = 0$$

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.

$$x^{2}+x-56=0$$

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.

$$x=7$$ $$x=-8$$