(x+3)(x+5)=7x+57

Solution for (x+3)(x+5)=7x+57 equation:

(x+3)(x+5)=7x+57

We move all terms to the left:

(x+3)(x+5)-(7x+57)=0

We get rid of parentheses

(x+3)(x+5)-7x-57=0

We multiply parentheses ..

(+x^2+5x+3x+15)-7x-57=0

We get rid of parentheses

x^2+5x+3x-7x+15-57=0

We add all the numbers together, and all the variables

x^2+x-42=0

a = 1; b = 1; c = -42;
Δ = b2-4ac
Δ = 12-4·1·(-42)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*1}=\frac{12}{2}
=6
$