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(v^2)+6v=11
We move all terms to the left:
(v^2)+6v-(11)=0
a = 1; b = 6; c = -11;
Δ = b2-4ac
Δ = 62-4·1·(-11)
Δ = 80
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{5}}{2*1}=\frac{-6-4\sqrt{5}}{2} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{5}}{2*1}=\frac{-6+4\sqrt{5}}{2} $
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