(X+2)(x+4)=57

Solution for (X+2)(x+4)=57 equation:

(X+2)(X+4)=57

We move all terms to the left:

(X+2)(X+4)-(57)=0

We multiply parentheses ..

(+X^2+4X+2X+8)-57=0

We get rid of parentheses

X^2+4X+2X+8-57=0

We add all the numbers together, and all the variables

X^2+6X-49=0

a = 1; b = 6; c = -49;
Δ = b2-4ac
Δ = 62-4·1·(-49)
Δ = 232
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{58}}{2*1}=\frac{-6-2\sqrt{58}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{58}}{2*1}=\frac{-6+2\sqrt{58}}{2}
$