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4q^2-23q+21=0
a = 4; b = -23; c = +21;
Δ = b2-4ac
Δ = -232-4·4·21
Δ = 193
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{-(-23)-\sqrt{193}}{2*4}=\frac{23-\sqrt{193}}{8} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{193}}{2*4}=\frac{23+\sqrt{193}}{8} $
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