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3x^2+60x+104=0
a = 3; b = 60; c = +104;
Δ = b2-4ac
Δ = 602-4·3·104
Δ = 2352
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-28\sqrt{3}}{2*3}=\frac{-60-28\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+28\sqrt{3}}{2*3}=\frac{-60+28\sqrt{3}}{6} $
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