# How do you factor the expression 10x^2+20x-80?

You factor a polynomial by finding (when possible) its roots ${x}_{i}$ and dividing the polynomial by the linear factor $\left(x - {x}_{i}\right)$.
In your case, first of all, let's factor $10$ out of the polynomial, obtaining
$10 {x}^{2} + 20 x - 80 = 10 \left({x}^{2} + 2 x - 8\right)$
Without bothering with the discriminant formula, we can solve ${x}^{2} + 2 x - 8$ by remembering the result that, when a quadratic polynomial is of the form ${x}^{2} - s x + p$, the sum of the solutions is $s$, and the product is $p$. In this case, we need to find two numbers ${x}_{1}$ and ${x}_{2}$ such that ${x}_{1} + {x}_{2} = - 2$, and ${x}_{1} {x}_{2} = - 8$.
It should be easy to find that the two numbers are $- 4$ and $2$. So, the linear factors are $\left(x + 4\right) \left(x - 2\right)$. The factorization of your polynomial is thus $10 \left(x + 4\right) \left(x - 2\right)$.