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25x^2-26=0
a = 25; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·25·(-26)
Δ = 2600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2600}=\sqrt{100*26}=\sqrt{100}*\sqrt{26}=10\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{26}}{2*25}=\frac{0-10\sqrt{26}}{50} =-\frac{10\sqrt{26}}{50} =-\frac{\sqrt{26}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{26}}{2*25}=\frac{0+10\sqrt{26}}{50} =\frac{10\sqrt{26}}{50} =\frac{\sqrt{26}}{5} $
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