(3x+4)(5x+1)+(3x+6)(5x+1)=0

Solution for (3x+4)(5x+1)+(3x+6)(5x+1)=0 equation:

(3x+4)(5x+1)+(3x+6)(5x+1)=0

We multiply parentheses ..

(+15x^2+3x+20x+4)+(3x+6)(5x+1)=0

We get rid of parentheses

15x^2+3x+20x+(3x+6)(5x+1)+4=0

We multiply parentheses ..

15x^2+(+15x^2+3x+30x+6)+3x+20x+4=0

We add all the numbers together, and all the variables

15x^2+(+15x^2+3x+30x+6)+23x+4=0

We get rid of parentheses

15x^2+15x^2+3x+30x+23x+6+4=0

We add all the numbers together, and all the variables

30x^2+56x+10=0

a = 30; b = 56; c = +10;
Δ = b2-4ac
Δ = 562-4·30·10
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-44}{2*30}=\frac{-100}{60}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+44}{2*30}=\frac{-12}{60}
=-1/5
$