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(5v+4)(5+v)=0
We add all the numbers together, and all the variables
(5v+4)(v+5)=0
We multiply parentheses ..
(+5v^2+25v+4v+20)=0
We get rid of parentheses
5v^2+25v+4v+20=0
We add all the numbers together, and all the variables
5v^2+29v+20=0
a = 5; b = 29; c = +20;
Δ = b2-4ac
Δ = 292-4·5·20
Δ = 441
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{441}=21$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*5}=\frac{-50}{10} =-5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*5}=\frac{-8}{10} =-4/5 $
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