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4v^2+16v-48=0

a = 4; b = 16; c = -48;

Δ = b^{2}-4ac

Δ = 16^{2}-4·4·(-48)

Δ = 1024

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{1024}=32$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*4}=\frac{-48}{8} =-6 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*4}=\frac{16}{8} =2 $

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