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x^2+1.5x-2.6=0
a = 1; b = 1.5; c = -2.6;
Δ = b2-4ac
Δ = 1.52-4·1·(-2.6)
Δ = 12.65
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{-(1.5)-\sqrt{12.65}}{2*1}=\frac{-1.5-\sqrt{12.65}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{12.65}}{2*1}=\frac{-1.5+\sqrt{12.65}}{2} $
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