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-4b^2+6b+28=0
a = -4; b = 6; c = +28;
Δ = b2-4ac
Δ = 62-4·(-4)·28
Δ = 484
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{484}=22$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-22}{2*-4}=\frac{-28}{-8} =3+1/2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+22}{2*-4}=\frac{16}{-8} =-2 $
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