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4x^2-19x+18=0
a = 4; b = -19; c = +18;
Δ = b2-4ac
Δ = -192-4·4·18
Δ = 73
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{73}}{2*4}=\frac{19-\sqrt{73}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{73}}{2*4}=\frac{19+\sqrt{73}}{8} $
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