(8x+3)(x+5)=0

Solution for (8x+3)(x+5)=0 equation:

(8x+3)(x+5)=0

We multiply parentheses ..

(+8x^2+40x+3x+15)=0

We get rid of parentheses

8x^2+40x+3x+15=0

We add all the numbers together, and all the variables

8x^2+43x+15=0

a = 8; b = 43; c = +15;
Δ = b2-4ac
Δ = 432-4·8·15
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-37}{2*8}=\frac{-80}{16}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+37}{2*8}=\frac{-6}{16}
=-3/8
$