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25x^2+22x-6=0
a = 25; b = 22; c = -6;
Δ = b2-4ac
Δ = 222-4·25·(-6)
Δ = 1084
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1084}=\sqrt{4*271}=\sqrt{4}*\sqrt{271}=2\sqrt{271}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{271}}{2*25}=\frac{-22-2\sqrt{271}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{271}}{2*25}=\frac{-22+2\sqrt{271}}{50} $
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