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391x^2+1298x+1015=0
a = 391; b = 1298; c = +1015;
Δ = b2-4ac
Δ = 12982-4·391·1015
Δ = 97344
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{97344}=312$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1298)-312}{2*391}=\frac{-1610}{782} =-2+1/17 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1298)+312}{2*391}=\frac{-986}{782} =-1+6/23 $
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