Quadratic Equation

$x^2+x-20=0$

By factoring, we have

$(x+5)(x-4)=0$

$x=-5$

$x=4$

So the positive root is $\boxed{x=4}$ .

Note: We can also use the quadratic formula.

x^2 +x - 20

this equals,

x^2 + 5x - 4x - 20

or,

x(x + 5) - 4(x + 5)

i.e.

(x - 4)(x + 5)

So, either x = 4 or x = -5

Hence positive root equals 4 !!!!!!

$x ^ { 2 } + x - 20 = 0$

$( x - 4 ) ( x + 5 ) = 0$

$x - 4 = 0 \text { , or } x + 5 = 0$ \

$x = 4 \text { , or } x = -5$

But, we are looking for the positive root, so it is $\boxed { \color{#3D99F6} x = 4 }$

One way to solve this is by factoring.

$x^2+x-20=0$

( $x+5)(x-4)=0$

$x+5=0$

$x=-5$

or

$x-4=0$

$x=4$

Therefore, the positive root is $x=4$ .