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7b^2+20b-32=0
a = 7; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·7·(-32)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-36}{2*7}=\frac{-56}{14} =-4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+36}{2*7}=\frac{16}{14} =1+1/7 $
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