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3x^2-12-5x=0
a = 3; b = -5; c = -12;
Δ = b2-4ac
Δ = -52-4·3·(-12)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*3}=\frac{-8}{6} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*3}=\frac{18}{6} =3 $
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