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q^2-4q-45=0
a = 1; b = -4; c = -45;
Δ = b2-4ac
Δ = -42-4·1·(-45)
Δ = 196
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{196}=14$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*1}=\frac{-10}{2} =-5 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*1}=\frac{18}{2} =9 $
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