(6v^2-3v-2)=(v^2-7v+5)

Solution for (6v^2-3v-2)=(v^2-7v+5) equation:

(6v^2-3v-2)=(v^2-7v+5)

We move all terms to the left:

(6v^2-3v-2)-((v^2-7v+5))=0

We get rid of parentheses

6v^2-3v-((v^2-7v+5))-2=0


We calculate terms in parentheses: -((v^2-7v+5)), so:

(v^2-7v+5)

We get rid of parentheses

v^2-7v+5

Back to the equation:

-(v^2-7v+5)


We get rid of parentheses

6v^2-v^2-3v+7v-5-2=0

We add all the numbers together, and all the variables

5v^2+4v-7=0

a = 5; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·5·(-7)
Δ = 156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{39}}{2*5}=\frac{-4-2\sqrt{39}}{10}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{39}}{2*5}=\frac{-4+2\sqrt{39}}{10}
$