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25x^2-5x-36=0
a = 25; b = -5; c = -36;
Δ = b2-4ac
Δ = -52-4·25·(-36)
Δ = 3625
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3625}=\sqrt{25*145}=\sqrt{25}*\sqrt{145}=5\sqrt{145}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{145}}{2*25}=\frac{5-5\sqrt{145}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{145}}{2*25}=\frac{5+5\sqrt{145}}{50} $
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