(14+x)(18+x)=437

Solution for (14+x)(18+x)=437 equation:

(14+x)(18+x)=437

We move all terms to the left:

(14+x)(18+x)-(437)=0

We add all the numbers together, and all the variables

(x+14)(x+18)-437=0

We multiply parentheses ..

(+x^2+18x+14x+252)-437=0

We get rid of parentheses

x^2+18x+14x+252-437=0

We add all the numbers together, and all the variables

x^2+32x-185=0

a = 1; b = 32; c = -185;
Δ = b2-4ac
Δ = 322-4·1·(-185)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-42}{2*1}=\frac{-74}{2}
=-37
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+42}{2*1}=\frac{10}{2}
=5
$