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4q^2-43q-11=0
a = 4; b = -43; c = -11;
Δ = b2-4ac
Δ = -432-4·4·(-11)
Δ = 2025
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{2025}=45$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-45}{2*4}=\frac{-2}{8} =-1/4 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+45}{2*4}=\frac{88}{8} =11 $
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