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x^2+3.5x-3=0
a = 1; b = 3.5; c = -3;
Δ = b2-4ac
Δ = 3.52-4·1·(-3)
Δ = 24.25
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{-(3.5)-\sqrt{24.25}}{2*1}=\frac{-3.5-\sqrt{24.25}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{24.25}}{2*1}=\frac{-3.5+\sqrt{24.25}}{2} $
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