(6x+5)(7x+1)=0

Solution for (6x+5)(7x+1)=0 equation:

(6x+5)(7x+1)=0

We multiply parentheses ..

(+42x^2+6x+35x+5)=0

We get rid of parentheses

42x^2+6x+35x+5=0

We add all the numbers together, and all the variables

42x^2+41x+5=0

a = 42; b = 41; c = +5;
Δ = b2-4ac
Δ = 412-4·42·5
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-29}{2*42}=\frac{-70}{84}
=-5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+29}{2*42}=\frac{-12}{84}
=-1/7
$