(7a+3)(a-4)=0

Solution for (7a+3)(a-4)=0 equation:

(7a+3)(a-4)=0

We multiply parentheses ..

(+7a^2-28a+3a-12)=0

We get rid of parentheses

7a^2-28a+3a-12=0

We add all the numbers together, and all the variables

7a^2-25a-12=0

a = 7; b = -25; c = -12;
Δ = b2-4ac
Δ = -252-4·7·(-12)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-31}{2*7}=\frac{-6}{14}
=-3/7
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+31}{2*7}=\frac{56}{14}
=4
$