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

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

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

We multiply parentheses ..

(+20x^2+15x+8x+6)=0

We get rid of parentheses

20x^2+15x+8x+6=0

We add all the numbers together, and all the variables

20x^2+23x+6=0

a = 20; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·20·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*20}=\frac{-30}{40}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*20}=\frac{-16}{40}
=-2/5
$