(x+3)(x+5)=48

Solution for (x+3)(x+5)=48 equation:

(x+3)(x+5)=48

We move all terms to the left:

(x+3)(x+5)-(48)=0

We multiply parentheses ..

(+x^2+5x+3x+15)-48=0

We get rid of parentheses

x^2+5x+3x+15-48=0

We add all the numbers together, and all the variables

x^2+8x-33=0

a = 1; b = 8; c = -33;
Δ = b2-4ac
Δ = 82-4·1·(-33)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-14}{2*1}=\frac{-22}{2}
=-11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+14}{2*1}=\frac{6}{2}
=3
$