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48x^2-49x+12=0
a = 48; b = -49; c = +12;
Δ = b2-4ac
Δ = -492-4·48·12
Δ = 97
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{-(-49)-\sqrt{97}}{2*48}=\frac{49-\sqrt{97}}{96} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{97}}{2*48}=\frac{49+\sqrt{97}}{96} $
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