-6v(v+9)=-9v-42

Solution for -6v(v+9)=-9v-42 equation:

-6v(v+9)=-9v-42

We move all terms to the left:

-6v(v+9)-(-9v-42)=0

We multiply parentheses

-6v^2-54v-(-9v-42)=0

We get rid of parentheses

-6v^2-54v+9v+42=0

We add all the numbers together, and all the variables

-6v^2-45v+42=0

a = -6; b = -45; c = +42;
Δ = b2-4ac
Δ = -452-4·(-6)·42
Δ = 3033
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:

$\sqrt{\Delta}=\sqrt{3033}=\sqrt{9*337}=\sqrt{9}*\sqrt{337}=3\sqrt{337}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-3\sqrt{337}}{2*-6}=\frac{45-3\sqrt{337}}{-12}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+3\sqrt{337}}{2*-6}=\frac{45+3\sqrt{337}}{-12}
$