# How do you solve x^2 +4x=21?

Mar 26, 2016

We are given that ${x}^{2} + 4 x = 21$, or ${x}^{2} + 4 x - 21 = 0$. This is a quadratic equation, which is in principle soluble. Factorization presents an easier method.
$\left(x + 7\right) \left(x - 3\right)$ $=$ ${x}^{2} + 4 x - 21$ upon expansion.
So, if $\left(x + 7\right) \left(x - 3\right)$ $=$ $0$, then clearly $x$ has roots of $- 7$ OR $+ 3$. Substitution of either of these values for $x$ gives $0$. Other posters may give a more systematic method for solution.