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4b^2+32b=57
We move all terms to the left:
4b^2+32b-(57)=0
a = 4; b = 32; c = -57;
Δ = b2-4ac
Δ = 322-4·4·(-57)
Δ = 1936
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{1936}=44$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-44}{2*4}=\frac{-76}{8} =-9+1/2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+44}{2*4}=\frac{12}{8} =1+1/2 $
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