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3b^2+16b-44=0
a = 3; b = 16; c = -44;
Δ = b2-4ac
Δ = 162-4·3·(-44)
Δ = 784
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{784}=28$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-28}{2*3}=\frac{-44}{6} =-7+1/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+28}{2*3}=\frac{12}{6} =2 $
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