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(2v-5)(5v+3)=0
We multiply parentheses ..
(+10v^2+6v-25v-15)=0
We get rid of parentheses
10v^2+6v-25v-15=0
We add all the numbers together, and all the variables
10v^2-19v-15=0
a = 10; b = -19; c = -15;
Δ = b2-4ac
Δ = -192-4·10·(-15)
Δ = 961
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{961}=31$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-31}{2*10}=\frac{-12}{20} =-3/5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+31}{2*10}=\frac{50}{20} =2+1/2 $
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