(x+5)(3x+4)-3(x+2)+4=0

Solution for (x+5)(3x+4)-3(x+2)+4=0 equation:

(x+5)(3x+4)-3(x+2)+4=0

We multiply parentheses

(x+5)(3x+4)-3x-6+4=0

We multiply parentheses ..

(+3x^2+4x+15x+20)-3x-6+4=0

We add all the numbers together, and all the variables

(+3x^2+4x+15x+20)-3x-2=0

We get rid of parentheses

3x^2+4x+15x-3x+20-2=0

We add all the numbers together, and all the variables

3x^2+16x+18=0

a = 3; b = 16; c = +18;
Δ = b2-4ac
Δ = 162-4·3·18
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{10}}{2*3}=\frac{-16-2\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{10}}{2*3}=\frac{-16+2\sqrt{10}}{6}
$