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2a^2+33a+136=0
a = 2; b = 33; c = +136;
Δ = b2-4ac
Δ = 332-4·2·136
Δ = 1
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{1}=1$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-1}{2*2}=\frac{-34}{4} =-8+1/2 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+1}{2*2}=\frac{-32}{4} =-8 $
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