(4v-3)(1-v)=0

Solution for (4v-3)(1-v)=0 equation:

(4v-3)(1-v)=0

We add all the numbers together, and all the variables

(4v-3)(-1v+1)=0

We multiply parentheses ..

(-4v^2+4v+3v-3)=0

We get rid of parentheses

-4v^2+4v+3v-3=0

We add all the numbers together, and all the variables

-4v^2+7v-3=0

a = -4; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·(-4)·(-3)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*-4}=\frac{-8}{-8}
=1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*-4}=\frac{-6}{-8}
=3/4
$