(4+x)(6+x)=48

Solution for (4+x)(6+x)=48 equation:

(4+x)(6+x)=48

We move all terms to the left:

(4+x)(6+x)-(48)=0

We add all the numbers together, and all the variables

(x+4)(x+6)-48=0

We multiply parentheses ..

(+x^2+6x+4x+24)-48=0

We get rid of parentheses

x^2+6x+4x+24-48=0

We add all the numbers together, and all the variables

x^2+10x-24=0

a = 1; b = 10; c = -24;
Δ = b2-4ac
Δ = 102-4·1·(-24)
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{-(10)-14}{2*1}=\frac{-24}{2}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*1}=\frac{4}{2}
=2
$