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x^2+1.5x=126
We move all terms to the left:
x^2+1.5x-(126)=0
a = 1; b = 1.5; c = -126;
Δ = b2-4ac
Δ = 1.52-4·1·(-126)
Δ = 506.25
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{-(1.5)-\sqrt{506.25}}{2*1}=\frac{-1.5-\sqrt{506.25}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{506.25}}{2*1}=\frac{-1.5+\sqrt{506.25}}{2} $
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