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-0.3x^2+3x+6=-6
We move all terms to the left:
-0.3x^2+3x+6-(-6)=0
We add all the numbers together, and all the variables
-0.3x^2+3x+12=0
a = -0.3; b = 3; c = +12;
Δ = b2-4ac
Δ = 32-4·(-0.3)·12
Δ = 23.4
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{-(3)-\sqrt{23.4}}{2*-0.3}=\frac{-3-\sqrt{23.4}}{-0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{23.4}}{2*-0.3}=\frac{-3+\sqrt{23.4}}{-0.6} $
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