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10x^2-3x-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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-27}{2*10}=\frac{-24}{20} =-1+1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+27}{2*10}=\frac{30}{20} =1+1/2 $
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