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3x^2-65x+348=0
a = 3; b = -65; c = +348;
Δ = b2-4ac
Δ = -652-4·3·348
Δ = 49
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-7}{2*3}=\frac{58}{6} =9+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+7}{2*3}=\frac{72}{6} =12 $
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