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8k^2+25k+3=0
a = 8; b = 25; c = +3;
Δ = b2-4ac
Δ = 252-4·8·3
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{529}=23$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*8}=\frac{-48}{16} =-3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*8}=\frac{-2}{16} =-1/8 $
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