(x-3)(x+5)=42

Solution for (x-3)(x+5)=42 equation:

(x-3)(x+5)=42

We move all terms to the left:

(x-3)(x+5)-(42)=0

We multiply parentheses ..

(+x^2+5x-3x-15)-42=0

We get rid of parentheses

x^2+5x-3x-15-42=0

We add all the numbers together, and all the variables

x^2+2x-57=0

a = 1; b = 2; c = -57;
Δ = b2-4ac
Δ = 22-4·1·(-57)
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{-(2)-2\sqrt{58}}{2*1}=\frac{-2-2\sqrt{58}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{58}}{2*1}=\frac{-2+2\sqrt{58}}{2}
$