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X^2+13X-35.5=0
a = 1; b = 13; c = -35.5;
Δ = b2-4ac
Δ = 132-4·1·(-35.5)
Δ = 311
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{-(13)-\sqrt{311}}{2*1}=\frac{-13-\sqrt{311}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{311}}{2*1}=\frac{-13+\sqrt{311}}{2} $
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