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12-26x+10x^2=0
a = 10; b = -26; c = +12;
Δ = b2-4ac
Δ = -262-4·10·12
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-14}{2*10}=\frac{12}{20} =3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+14}{2*10}=\frac{40}{20} =2 $
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