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

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

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

We multiply parentheses

3x^2+6x+12x-20=0

We add all the numbers together, and all the variables

3x^2+18x-20=0

a = 3; b = 18; c = -20;
Δ = b2-4ac
Δ = 182-4·3·(-20)
Δ = 564
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:

$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{141}}{2*3}=\frac{-18-2\sqrt{141}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{141}}{2*3}=\frac{-18+2\sqrt{141}}{6}
$