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2v^2-5=23
We move all terms to the left:
2v^2-5-(23)=0
We add all the numbers together, and all the variables
2v^2-28=0
a = 2; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·2·(-28)
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{-(0)-4\sqrt{14}}{2*2}=\frac{0-4\sqrt{14}}{4} =-\frac{4\sqrt{14}}{4} =-\sqrt{14} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{14}}{2*2}=\frac{0+4\sqrt{14}}{4} =\frac{4\sqrt{14}}{4} =\sqrt{14} $
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