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3x^2-38x+88=0
a = 3; b = -38; c = +88;
Δ = b2-4ac
Δ = -382-4·3·88
Δ = 388
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{97}}{2*3}=\frac{38-2\sqrt{97}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{97}}{2*3}=\frac{38+2\sqrt{97}}{6} $
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