(X+5)(x+5)=4x+41

Solution for (X+5)(x+5)=4x+41 equation:

(X+5)(X+5)=4X+41

We move all terms to the left:

(X+5)(X+5)-(4X+41)=0

We get rid of parentheses

(X+5)(X+5)-4X-41=0

We multiply parentheses ..

(+X^2+5X+5X+25)-4X-41=0

We get rid of parentheses

X^2+5X+5X-4X+25-41=0

We add all the numbers together, and all the variables

X^2+6X-16=0

a = 1; b = 6; c = -16;
Δ = b2-4ac
Δ = 62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*1}=\frac{-16}{2}
=-8
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*1}=\frac{4}{2}
=2
$