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4b^2-4b-35=0
a = 4; b = -4; c = -35;
Δ = b2-4ac
Δ = -42-4·4·(-35)
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$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-24}{2*4}=\frac{-20}{8} =-2+1/2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+24}{2*4}=\frac{28}{8} =3+1/2 $
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