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4b^2-5b-21=0
a = 4; b = -5; c = -21;
Δ = b2-4ac
Δ = -52-4·4·(-21)
Δ = 361
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{361}=19$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*4}=\frac{-14}{8} =-1+3/4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*4}=\frac{24}{8} =3 $
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