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4v^2=12v+5
We move all terms to the left:
4v^2-(12v+5)=0
We get rid of parentheses
4v^2-12v-5=0
a = 4; b = -12; c = -5;
Δ = b2-4ac
Δ = -122-4·4·(-5)
Δ = 224
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{14}}{2*4}=\frac{12-4\sqrt{14}}{8} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{14}}{2*4}=\frac{12+4\sqrt{14}}{8} $
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