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18a^2+3a-10=0
a = 18; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·18·(-10)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{729}=27$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*18}=\frac{-30}{36} =-5/6 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*18}=\frac{24}{36} =2/3 $
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