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3x^2-26x+56=0
a = 3; b = -26; c = +56;
Δ = b2-4ac
Δ = -262-4·3·56
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2}{2*3}=\frac{24}{6} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2}{2*3}=\frac{28}{6} =4+2/3 $
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