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2b^2+9b-18=0
a = 2; b = 9; c = -18;
Δ = b2-4ac
Δ = 92-4·2·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{225}=15$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*2}=\frac{-24}{4} =-6 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*2}=\frac{6}{4} =1+1/2 $
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