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10x^2-38x+28=0
a = 10; b = -38; c = +28;
Δ = b2-4ac
Δ = -382-4·10·28
Δ = 324
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-18}{2*10}=\frac{20}{20} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+18}{2*10}=\frac{56}{20} =2+4/5 $
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