2v^2+9=-9v

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Solution for 2v^2+9=-9v equation:



2v^2+9=-9v
We move all terms to the left:
2v^2+9-(-9v)=0
We get rid of parentheses
2v^2+9v+9=0
a = 2; b = 9; c = +9;
Δ = b2-4ac
Δ = 92-4·2·9
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{-(9)-3}{2*2}=\frac{-12}{4} =-3 $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*2}=\frac{-6}{4} =-1+1/2 $

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