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25z^2-45=0
a = 25; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·25·(-45)
Δ = 4500
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4500}=\sqrt{900*5}=\sqrt{900}*\sqrt{5}=30\sqrt{5}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{5}}{2*25}=\frac{0-30\sqrt{5}}{50} =-\frac{30\sqrt{5}}{50} =-\frac{3\sqrt{5}}{5} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{5}}{2*25}=\frac{0+30\sqrt{5}}{50} =\frac{30\sqrt{5}}{50} =\frac{3\sqrt{5}}{5} $
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