2v(v-7)=5v(v-7)

Solution for 2v(v-7)=5v(v-7) equation:

2v(v-7)=5v(v-7)

We move all terms to the left:

2v(v-7)-(5v(v-7))=0

We multiply parentheses

2v^2-14v-(5v(v-7))=0


We calculate terms in parentheses: -(5v(v-7)), so:

5v(v-7)

We multiply parentheses

5v^2-35v

Back to the equation:

-(5v^2-35v)


We get rid of parentheses

2v^2-5v^2-14v+35v=0

We add all the numbers together, and all the variables

-3v^2+21v=0

a = -3; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·(-3)·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*-3}=\frac{-42}{-6}
=+7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*-3}=\frac{0}{-6}
=0
$