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(6v-5)(v-4)=0
We multiply parentheses ..
(+6v^2-24v-5v+20)=0
We get rid of parentheses
6v^2-24v-5v+20=0
We add all the numbers together, and all the variables
6v^2-29v+20=0
a = 6; b = -29; c = +20;
Δ = b2-4ac
Δ = -292-4·6·20
Δ = 361
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{361}=19$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-19}{2*6}=\frac{10}{12} =5/6 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+19}{2*6}=\frac{48}{12} =4 $
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