Question

Solve x^2-3x-20=8 for quadratic equation​

Answer 1

We have

$x^2-3x-20=8$

First, move $8$ to left side to get

$x^2-3x-28=0$

Rewrite $x^2-3x-28$ as $(x-7)(x+4)$ to get

$(x-7)(x+4)=0$

The expression on the left equals zero if either of factors is zero thus the equation has 2 solutions

$x-7=0\Longrightarrow\boxed{x_1=7}$

$x+4=0\Longrightarrow\boxed{x_2=-4}$

Hope this helps.

Answer 2

Answer:

x = -7

x = 4

Step-by-step explanation: