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3x^2+4x+8=-6x
We move all terms to the left:
3x^2+4x+8-(-6x)=0
We get rid of parentheses
3x^2+4x+6x+8=0
We add all the numbers together, and all the variables
3x^2+10x+8=0
a = 3; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·3·8
Δ = 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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*3}=\frac{-12}{6} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*3}=\frac{-8}{6} =-1+1/3 $
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