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3x^2-50x+126=0
a = 3; b = -50; c = +126;
Δ = b2-4ac
Δ = -502-4·3·126
Δ = 988
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{988}=\sqrt{4*247}=\sqrt{4}*\sqrt{247}=2\sqrt{247}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{247}}{2*3}=\frac{50-2\sqrt{247}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{247}}{2*3}=\frac{50+2\sqrt{247}}{6} $
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