x^2+x+4x+4=0

Solution for x^2+x+4x+4=0 equation:

x^2+x+4x+4=0

We add all the numbers together, and all the variables

x^2+5x+4=0

a = 1; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·1·4
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*1}=\frac{-2}{2}
=-1
$