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3x^2+x-10=0
a = 3; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·3·(-10)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*3}=\frac{-12}{6} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*3}=\frac{10}{6} =1+2/3 $
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