(7x+1)(3x+4)=0

Solution for (7x+1)(3x+4)=0 equation:

(7x+1)(3x+4)=0

We multiply parentheses ..

(+21x^2+28x+3x+4)=0

We get rid of parentheses

21x^2+28x+3x+4=0

We add all the numbers together, and all the variables

21x^2+31x+4=0

a = 21; b = 31; c = +4;
Δ = b2-4ac
Δ = 312-4·21·4
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-25}{2*21}=\frac{-56}{42}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+25}{2*21}=\frac{-6}{42}
=-1/7
$