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(q^2)+(346.9295q)+2642.5133=0
We add all the numbers together, and all the variables
q^2+(+346.9295q)+2642.5133=0
We get rid of parentheses
q^2+346.9295q+2642.5133=0
a = 1; b = 346.9295; c = +2642.5133;
Δ = b2-4ac
Δ = 346.92952-4·1·2642.5133
Δ = 109790.02477
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{-(346.9295)-\sqrt{109790.02477}}{2*1}=\frac{-346.9295-\sqrt{109790.02477}}{2} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(346.9295)+\sqrt{109790.02477}}{2*1}=\frac{-346.9295+\sqrt{109790.02477}}{2} $
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