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3x^2+5-13x=0
a = 3; b = -13; c = +5;
Δ = b2-4ac
Δ = -132-4·3·5
Δ = 109
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{-(-13)-\sqrt{109}}{2*3}=\frac{13-\sqrt{109}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{109}}{2*3}=\frac{13+\sqrt{109}}{6} $
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