(4v-3)(v-5=)

Solution for (4v-3)(v-5=) equation:

(4v-3)(v-5=)

We move all terms to the left:

(4v-3)(v-5-())=0


We calculate terms in parentheses: +(4v-3)(v-5-()), so:

4v-3)(v-5-()

We add all the numbers together, and all the variables

4v-3)(v

Back to the equation:

+(4v-3)(v)


We multiply parentheses

4v^2-3v=0

a = 4; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·4·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*4}=\frac{0}{8}
=0
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*4}=\frac{6}{8}
=3/4
$