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25x^2-20x-54=0
a = 25; b = -20; c = -54;
Δ = b2-4ac
Δ = -202-4·25·(-54)
Δ = 5800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5800}=\sqrt{100*58}=\sqrt{100}*\sqrt{58}=10\sqrt{58}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{58}}{2*25}=\frac{20-10\sqrt{58}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{58}}{2*25}=\frac{20+10\sqrt{58}}{50} $
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