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3x^2-14x+10=0
a = 3; b = -14; c = +10;
Δ = b2-4ac
Δ = -142-4·3·10
Δ = 76
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{19}}{2*3}=\frac{14-2\sqrt{19}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{19}}{2*3}=\frac{14+2\sqrt{19}}{6} $
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