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=5-4Y^2
We move all terms to the left:
-(5-4Y^2)=0
We get rid of parentheses
4Y^2-5=0
a = 4; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·4·(-5)
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*4}=\frac{0-4\sqrt{5}}{8} =-\frac{4\sqrt{5}}{8} =-\frac{\sqrt{5}}{2} $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*4}=\frac{0+4\sqrt{5}}{8} =\frac{4\sqrt{5}}{8} =\frac{\sqrt{5}}{2} $
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