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20q^2+56q+36=0
a = 20; b = 56; c = +36;
Δ = b2-4ac
Δ = 562-4·20·36
Δ = 256
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{256}=16$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-16}{2*20}=\frac{-72}{40} =-1+4/5 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+16}{2*20}=\frac{-40}{40} =-1 $
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