(3f+7)(4-2f)=0

Solution for (3f+7)(4-2f)=0 equation:

(3f+7)(4-2f)=0

We add all the numbers together, and all the variables

(3f+7)(-2f+4)=0

We multiply parentheses ..

(-6f^2+12f-14f+28)=0

We get rid of parentheses

-6f^2+12f-14f+28=0

We add all the numbers together, and all the variables

-6f^2-2f+28=0

a = -6; b = -2; c = +28;
Δ = b2-4ac
Δ = -22-4·(-6)·28
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-26}{2*-6}=\frac{-24}{-12}
=+2
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+26}{2*-6}=\frac{28}{-12}
=-2+1/3
$