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25x^2-150x+205=0
a = 25; b = -150; c = +205;
Δ = b2-4ac
Δ = -1502-4·25·205
Δ = 2000
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-150)-20\sqrt{5}}{2*25}=\frac{150-20\sqrt{5}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-150)+20\sqrt{5}}{2*25}=\frac{150+20\sqrt{5}}{50} $
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