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7x^2-19x-6=0
a = 7; b = -19; c = -6;
Δ = b2-4ac
Δ = -192-4·7·(-6)
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-23}{2*7}=\frac{-4}{14} =-2/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+23}{2*7}=\frac{42}{14} =3 $
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