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.3x^2=-1.2x
We move all terms to the left:
.3x^2-(-1.2x)=0
We get rid of parentheses
.3x^2+1.2x=0
a = .3; b = 1.2; c = 0;
Δ = b2-4ac
Δ = 1.22-4·.3·0
Δ = 1.44
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{-(1.2)-\sqrt{1.44}}{2*.3}=\frac{-1.2-\sqrt{1.44}}{0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{1.44}}{2*.3}=\frac{-1.2+\sqrt{1.44}}{0.6} $
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