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-(v+5)(5v+3)=0
We multiply parentheses ..
-(+5v^2+3v+25v+15)=0
We get rid of parentheses
-5v^2-3v-25v-15=0
We add all the numbers together, and all the variables
-5v^2-28v-15=0
a = -5; b = -28; c = -15;
Δ = b2-4ac
Δ = -282-4·(-5)·(-15)
Δ = 484
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{484}=22$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-22}{2*-5}=\frac{6}{-10} =-3/5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+22}{2*-5}=\frac{50}{-10} =-5 $
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