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x^2+23x-1058=0
a = 1; b = 23; c = -1058;
Δ = b2-4ac
Δ = 232-4·1·(-1058)
Δ = 4761
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{4761}=69$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-69}{2*1}=\frac{-92}{2} =-46 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+69}{2*1}=\frac{46}{2} =23 $
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