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v2+48v=0
We add all the numbers together, and all the variables
v^2+48v=0
a = 1; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·1·0
Δ = 2304
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{2304}=48$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*1}=\frac{-96}{2} =-48 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*1}=\frac{0}{2} =0 $
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