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4x^2+19x=111
We move all terms to the left:
4x^2+19x-(111)=0
a = 4; b = 19; c = -111;
Δ = b2-4ac
Δ = 192-4·4·(-111)
Δ = 2137
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{2137}}{2*4}=\frac{-19-\sqrt{2137}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{2137}}{2*4}=\frac{-19+\sqrt{2137}}{8} $
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