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(q^2)+(346.9318q)-2745.8607=0
We add all the numbers together, and all the variables
q^2+(+346.9318q)-2745.8607=0
We get rid of parentheses
q^2+346.9318q-2745.8607=0
a = 1; b = 346.9318; c = -2745.8607;
Δ = b2-4ac
Δ = 346.93182-4·1·(-2745.8607)
Δ = 131345.116651
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{-(346.9318)-\sqrt{131345.116651}}{2*1}=\frac{-346.9318-\sqrt{131345.116651}}{2} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(346.9318)+\sqrt{131345.116651}}{2*1}=\frac{-346.9318+\sqrt{131345.116651}}{2} $
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