(7x-5)(x+4)=0

Solution for (7x-5)(x+4)=0 equation:

(7x-5)(x+4)=0

We multiply parentheses ..

(+7x^2+28x-5x-20)=0

We get rid of parentheses

7x^2+28x-5x-20=0

We add all the numbers together, and all the variables

7x^2+23x-20=0

a = 7; b = 23; c = -20;
Δ = b2-4ac
Δ = 232-4·7·(-20)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-33}{2*7}=\frac{-56}{14}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+33}{2*7}=\frac{10}{14}
=5/7
$