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4c^2+19c-6=0
a = 4; b = 19; c = -6;
Δ = b2-4ac
Δ = 192-4·4·(-6)
Δ = 457
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{457}}{2*4}=\frac{-19-\sqrt{457}}{8} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{457}}{2*4}=\frac{-19+\sqrt{457}}{8} $
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