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x^2+0.15x=150
We move all terms to the left:
x^2+0.15x-(150)=0
a = 1; b = 0.15; c = -150;
Δ = b2-4ac
Δ = 0.152-4·1·(-150)
Δ = 600.0225
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{-(0.15)-\sqrt{600.0225}}{2*1}=\frac{-0.15-\sqrt{600.0225}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.15)+\sqrt{600.0225}}{2*1}=\frac{-0.15+\sqrt{600.0225}}{2} $
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