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2a^2+9a-11=0
a = 2; b = 9; c = -11;
Δ = b2-4ac
Δ = 92-4·2·(-11)
Δ = 169
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{169}=13$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-13}{2*2}=\frac{-22}{4} =-5+1/2 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+13}{2*2}=\frac{4}{4} =1 $
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