(x+3)(x+4)-20=0

Solution for (x+3)(x+4)-20=0 equation:

(x+3)(x+4)-20=0

We multiply parentheses ..

(+x^2+4x+3x+12)-20=0

We get rid of parentheses

x^2+4x+3x+12-20=0

We add all the numbers together, and all the variables

x^2+7x-8=0

a = 1; b = 7; c = -8;
Δ = b2-4ac
Δ = 72-4·1·(-8)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*1}=\frac{2}{2}
=1
$