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8k^2+2k-3=0
a = 8; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·8·(-3)
Δ = 100
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{100}=10$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*8}=\frac{-12}{16} =-3/4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*8}=\frac{8}{16} =1/2 $
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