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6x^2-35x+48=0
a = 6; b = -35; c = +48;
Δ = b2-4ac
Δ = -352-4·6·48
Δ = 73
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{73}}{2*6}=\frac{35-\sqrt{73}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{73}}{2*6}=\frac{35+\sqrt{73}}{12} $
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