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2x^2-57x+148=0
a = 2; b = -57; c = +148;
Δ = b2-4ac
Δ = -572-4·2·148
Δ = 2065
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{-(-57)-\sqrt{2065}}{2*2}=\frac{57-\sqrt{2065}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-57)+\sqrt{2065}}{2*2}=\frac{57+\sqrt{2065}}{4} $
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