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4x^2-19x+12=0
a = 4; b = -19; c = +12;
Δ = b2-4ac
Δ = -192-4·4·12
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*4}=\frac{6}{8} =3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*4}=\frac{32}{8} =4 $
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