(x+1)(x+4)=8

Solution for (x+1)(x+4)=8 equation:

(x+1)(x+4)=8

We move all terms to the left:

(x+1)(x+4)-(8)=0

We multiply parentheses ..

(+x^2+4x+x+4)-8=0

We get rid of parentheses

x^2+4x+x+4-8=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 = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{41}}{2*1}=\frac{-5-\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{41}}{2*1}=\frac{-5+\sqrt{41}}{2}
$