0=(7x+5)(2x+3)

Solution for 0=(7x+5)(2x+3) equation:

0=(7x+5)(2x+3)

We move all terms to the left:

0-((7x+5)(2x+3))=0

We add all the numbers together, and all the variables

-((7x+5)(2x+3))=0

We multiply parentheses ..

-((+14x^2+21x+10x+15))=0


We calculate terms in parentheses: -((+14x^2+21x+10x+15)), so:

(+14x^2+21x+10x+15)

We get rid of parentheses

14x^2+21x+10x+15

We add all the numbers together, and all the variables

14x^2+31x+15

Back to the equation:

-(14x^2+31x+15)


We get rid of parentheses

-14x^2-31x-15=0

a = -14; b = -31; c = -15;
Δ = b2-4ac
Δ = -312-4·(-14)·(-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{-(-31)-11}{2*-14}=\frac{20}{-28}
=-5/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+11}{2*-14}=\frac{42}{-28}
=-1+1/2
$