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(4v+3)(5+v)=0
We add all the numbers together, and all the variables
(4v+3)(v+5)=0
We multiply parentheses ..
(+4v^2+20v+3v+15)=0
We get rid of parentheses
4v^2+20v+3v+15=0
We add all the numbers together, and all the variables
4v^2+23v+15=0
a = 4; b = 23; c = +15;
Δ = b2-4ac
Δ = 232-4·4·15
Δ = 289
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{289}=17$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*4}=\frac{-40}{8} =-5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*4}=\frac{-6}{8} =-3/4 $
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