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3x^2+x-2=0
a = 3; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·3·(-2)
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*3}=\frac{-6}{6} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*3}=\frac{4}{6} =2/3 $
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