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26x^2-47x+12=0
a = 26; b = -47; c = +12;
Δ = b2-4ac
Δ = -472-4·26·12
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-31}{2*26}=\frac{16}{52} =4/13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+31}{2*26}=\frac{78}{52} =1+1/2 $
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