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4x^2-19x-23=0
a = 4; b = -19; c = -23;
Δ = b2-4ac
Δ = -192-4·4·(-23)
Δ = 729
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-27}{2*4}=\frac{-8}{8} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+27}{2*4}=\frac{46}{8} =5+3/4 $
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