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4x^2-19x-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:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{457}}{2*4}=\frac{19-\sqrt{457}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{457}}{2*4}=\frac{19+\sqrt{457}}{8} $
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