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3x^2+6x=-5x-6
We move all terms to the left:
3x^2+6x-(-5x-6)=0
We get rid of parentheses
3x^2+6x+5x+6=0
We add all the numbers together, and all the variables
3x^2+11x+6=0
a = 3; b = 11; c = +6;
Δ = b2-4ac
Δ = 112-4·3·6
Δ = 49
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*3}=\frac{-18}{6} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*3}=\frac{-4}{6} =-2/3 $
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