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x^2+0.2x=60
We move all terms to the left:
x^2+0.2x-(60)=0
a = 1; b = 0.2; c = -60;
Δ = b2-4ac
Δ = 0.22-4·1·(-60)
Δ = 240.04
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{-(0.2)-\sqrt{240.04}}{2*1}=\frac{-0.2-\sqrt{240.04}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{240.04}}{2*1}=\frac{-0.2+\sqrt{240.04}}{2} $
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