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