(x+3)x+4=20

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

(x+3)x+4=20

We move all terms to the left:

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

We add all the numbers together, and all the variables

(x+3)x-16=0

We multiply parentheses

x^2+3x-16=0

a = 1; b = 3; c = -16;
Δ = b2-4ac
Δ = 32-4·1·(-16)
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{73}}{2*1}=\frac{-3-\sqrt{73}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{73}}{2*1}=\frac{-3+\sqrt{73}}{2}
$