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12x^2-26x+10=0
a = 12; b = -26; c = +10;
Δ = b2-4ac
Δ = -262-4·12·10
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-14}{2*12}=\frac{12}{24} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+14}{2*12}=\frac{40}{24} =1+2/3 $
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