-v^2+6v-11=-3

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Solution for -v^2+6v-11=-3 equation:



-v^2+6v-11=-3
We move all terms to the left:
-v^2+6v-11-(-3)=0
We add all the numbers together, and all the variables
-1v^2+6v-8=0
a = -1; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·(-1)·(-8)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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

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