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q^2-21q+68=0
a = 1; b = -21; c = +68;
Δ = b2-4ac
Δ = -212-4·1·68
Δ = 169
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{169}=13$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-13}{2*1}=\frac{8}{2} =4 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+13}{2*1}=\frac{34}{2} =17 $
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