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(b-4)(3b+12)=0
We multiply parentheses ..
(+3b^2+12b-12b-48)=0
We get rid of parentheses
3b^2+12b-12b-48=0
We add all the numbers together, and all the variables
3b^2-48=0
a = 3; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·3·(-48)
Δ = 576
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{576}=24$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*3}=\frac{-24}{6} =-4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*3}=\frac{24}{6} =4 $
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