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v2=52
We move all terms to the left:
v2-(52)=0
We add all the numbers together, and all the variables
v^2-52=0
a = 1; b = 0; c = -52;
Δ = b2-4ac
Δ = 02-4·1·(-52)
Δ = 208
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{13}}{2*1}=\frac{0-4\sqrt{13}}{2} =-\frac{4\sqrt{13}}{2} =-2\sqrt{13} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{13}}{2*1}=\frac{0+4\sqrt{13}}{2} =\frac{4\sqrt{13}}{2} =2\sqrt{13} $
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