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2.5x^2+4x-254=0
a = 2.5; b = 4; c = -254;
Δ = b2-4ac
Δ = 42-4·2.5·(-254)
Δ = 2556
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2556}=\sqrt{36*71}=\sqrt{36}*\sqrt{71}=6\sqrt{71}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{71}}{2*2.5}=\frac{-4-6\sqrt{71}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{71}}{2*2.5}=\frac{-4+6\sqrt{71}}{5} $
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