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1.25x^2-10x+10=0
a = 1.25; b = -10; c = +10;
Δ = b2-4ac
Δ = -102-4·1.25·10
Δ = 50
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{50}=\sqrt{25*2}=\sqrt{25}*\sqrt{2}=5\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-5\sqrt{2}}{2*1.25}=\frac{10-5\sqrt{2}}{2.5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+5\sqrt{2}}{2*1.25}=\frac{10+5\sqrt{2}}{2.5} $
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