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3x^2-61x+24=0
a = 3; b = -61; c = +24;
Δ = b2-4ac
Δ = -612-4·3·24
Δ = 3433
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-61)-\sqrt{3433}}{2*3}=\frac{61-\sqrt{3433}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-61)+\sqrt{3433}}{2*3}=\frac{61+\sqrt{3433}}{6} $
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