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25x^2+40x-4=0
a = 25; b = 40; c = -4;
Δ = b2-4ac
Δ = 402-4·25·(-4)
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{-(40)-20\sqrt{5}}{2*25}=\frac{-40-20\sqrt{5}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{5}}{2*25}=\frac{-40+20\sqrt{5}}{50} $
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