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0=(v+1)(v+5)
We move all terms to the left:
0-((v+1)(v+5))=0
We add all the numbers together, and all the variables
-((v+1)(v+5))=0
We multiply parentheses ..
-((+v^2+5v+v+5))=0
We calculate terms in parentheses: -((+v^2+5v+v+5)), so:We get rid of parentheses
(+v^2+5v+v+5)
We get rid of parentheses
v^2+5v+v+5
We add all the numbers together, and all the variables
v^2+6v+5
Back to the equation:
-(v^2+6v+5)
-v^2-6v-5=0
We add all the numbers together, and all the variables
-1v^2-6v-5=0
a = -1; b = -6; c = -5;
Δ = b2-4ac
Δ = -62-4·(-1)·(-5)
Δ = 16
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{16}=4$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4}{2*-1}=\frac{2}{-2} =-1 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4}{2*-1}=\frac{10}{-2} =-5 $
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