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X^2+4.5X-130=0
a = 1; b = 4.5; c = -130;
Δ = b2-4ac
Δ = 4.52-4·1·(-130)
Δ = 540.25
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{-(4.5)-\sqrt{540.25}}{2*1}=\frac{-4.5-\sqrt{540.25}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.5)+\sqrt{540.25}}{2*1}=\frac{-4.5+\sqrt{540.25}}{2} $
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