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3x+10x^2-18=0
a = 10; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·10·(-18)
Δ = 729
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{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*10}=\frac{-30}{20} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*10}=\frac{24}{20} =1+1/5 $
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