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v^2=10
We move all terms to the left:
v^2-(10)=0
a = 1; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·1·(-10)
Δ = 40
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*1}=\frac{0-2\sqrt{10}}{2} =-\frac{2\sqrt{10}}{2} =-\sqrt{10} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*1}=\frac{0+2\sqrt{10}}{2} =\frac{2\sqrt{10}}{2} =\sqrt{10} $
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