0=(8x+5)(7x+3)

Solution for 0=(8x+5)(7x+3) equation:

0=(8x+5)(7x+3)

We move all terms to the left:

0-((8x+5)(7x+3))=0

We add all the numbers together, and all the variables

-((8x+5)(7x+3))=0

We multiply parentheses ..

-((+56x^2+24x+35x+15))=0


We calculate terms in parentheses: -((+56x^2+24x+35x+15)), so:

(+56x^2+24x+35x+15)

We get rid of parentheses

56x^2+24x+35x+15

We add all the numbers together, and all the variables

56x^2+59x+15

Back to the equation:

-(56x^2+59x+15)


We get rid of parentheses

-56x^2-59x-15=0

a = -56; b = -59; c = -15;
Δ = b2-4ac
Δ = -592-4·(-56)·(-15)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-59)-11}{2*-56}=\frac{48}{-112}
=-3/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-59)+11}{2*-56}=\frac{70}{-112}
=-5/8
$