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2q^2+31q+15=0
a = 2; b = 31; c = +15;
Δ = b2-4ac
Δ = 312-4·2·15
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{841}=29$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-29}{2*2}=\frac{-60}{4} =-15 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+29}{2*2}=\frac{-2}{4} =-1/2 $
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