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3x^2-13x-30=0
a = 3; b = -13; c = -30;
Δ = b2-4ac
Δ = -132-4·3·(-30)
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-23}{2*3}=\frac{-10}{6} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+23}{2*3}=\frac{36}{6} =6 $
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