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2y^2=10
We move all terms to the left:
2y^2-(10)=0
a = 2; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·2·(-10)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*2}=\frac{0-4\sqrt{5}}{4} =-\frac{4\sqrt{5}}{4} =-\sqrt{5} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*2}=\frac{0+4\sqrt{5}}{4} =\frac{4\sqrt{5}}{4} =\sqrt{5} $
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