1. ## Algebra prob

solve: $5-\sqrt{x+5}=2x+5$
i got $4x^2+x+5=0$ for a quadratic

but i believe it to be incorrect because the answer is x= -1 and i don't get that when i use the quad formula or complete the square so i think i messed up the quadratic

4x^2 - x - 5 = 0 ?

3. why is it $4x^2-x-5=0$ instead of $4x^2+x+5=0$

4. subtract 5 from both sides

-sqr(x+5) = 2x

square both sides

x+5 = 4x^2

take x + 5 from both sides

0 = 4x^2 -x -5

5. Originally Posted by ~berserk
why is it $4x^2-x-5=0$ instead of $4x^2+x+5=0$
$5-\sqrt{x+5} = 2x+5$

$-\sqrt{x+5} = 2x$

$(\sqrt{x+5})^2=(-2x)^2$

$x + 5 = 4x^2$

$4x^2 - x -5 =0$