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6b^2+25b+4=0
a = 6; b = 25; c = +4;
Δ = b2-4ac
Δ = 252-4·6·4
Δ = 529
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{529}=23$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*6}=\frac{-48}{12} =-4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*6}=\frac{-2}{12} =-1/6 $
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