(-3v2+6v-7)+(4v2+v+5)=0

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Solution for (-3v2+6v-7)+(4v2+v+5)=0 equation:



(-3v^2+6v-7)+(4v^2+v+5)=0
We get rid of parentheses
-3v^2+4v^2+6v+v-7+5=0
We add all the numbers together, and all the variables
v^2+7v-2=0
a = 1; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·1·(-2)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{57}}{2*1}=\frac{-7-\sqrt{57}}{2} $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{57}}{2*1}=\frac{-7+\sqrt{57}}{2} $

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