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53=v^2+3
We move all terms to the left:
53-(v^2+3)=0
We get rid of parentheses
-v^2-3+53=0
We add all the numbers together, and all the variables
-1v^2+50=0
a = -1; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-1)·50
Δ = 200
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*-1}=\frac{0-10\sqrt{2}}{-2} =-\frac{10\sqrt{2}}{-2} =-\frac{5\sqrt{2}}{-1} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*-1}=\frac{0+10\sqrt{2}}{-2} =\frac{10\sqrt{2}}{-2} =\frac{5\sqrt{2}}{-1} $
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