How would I solve the quadratic $x^2+3x-70=0$?

Question

How would I solve the following quadratic equation

$$x^2+3x-70=0 $$

This is my attempt below

$$(x-7x) (x+10x)=0 $$

$$ x-7x=0 \implies -6x=0 \implies x=6$$

$$x+10x=0 \implies 11x=0 \implies x=-11$$

Am I correct in thinking that the two solutions for $x$ are $x=6, x=-11$?

Answer 1

Your factoring is a bit off:

$$\begin{align}x^2 + 3x-70 = (x - 7)(x+10) = 0 & \iff (x - 7) = 0 \; \text{or}\; (x + 10) = 0 \\ \\ &\iff x = 7 \;\text{or} \;x = -10\end{align}$$

Instead of the above factors, you wrote $$(x - 7x)(x + 10x) = (-6x)(11x) = -66x^2 \neq x^2 + 3x - 70$$

Answer 2

Just one thing to add to amWhy's answer there's this formula for solving quadratic equations:

$$x_{1,2} = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where $a,b,c$ are the coefficients of $x^2, x,x^0$ respectively.

from which you get the two solutions ($x_1$ and $x_2$) for these kind of equations.