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3x^2-38x+95=0
a = 3; b = -38; c = +95;
Δ = b2-4ac
Δ = -382-4·3·95
Δ = 304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-4\sqrt{19}}{2*3}=\frac{38-4\sqrt{19}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+4\sqrt{19}}{2*3}=\frac{38+4\sqrt{19}}{6} $
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