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q2-0.3006q-0.0145=0
We add all the numbers together, and all the variables
q^2-0.3006q-0.0145=0
a = 1; b = -0.3006; c = -0.0145;
Δ = b2-4ac
Δ = -0.30062-4·1·(-0.0145)
Δ = 0.14836036
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.3006)-\sqrt{0.14836036}}{2*1}=\frac{0.3006-\sqrt{0.14836036}}{2} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.3006)+\sqrt{0.14836036}}{2*1}=\frac{0.3006+\sqrt{0.14836036}}{2} $
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