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10x^2-14x+4=0
a = 10; b = -14; c = +4;
Δ = b2-4ac
Δ = -142-4·10·4
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*10}=\frac{8}{20} =2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*10}=\frac{20}{20} =1 $
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