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-10x^2+25x-3=0
a = -10; b = 25; c = -3;
Δ = b2-4ac
Δ = 252-4·(-10)·(-3)
Δ = 505
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{505}}{2*-10}=\frac{-25-\sqrt{505}}{-20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{505}}{2*-10}=\frac{-25+\sqrt{505}}{-20} $
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