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x^2-x-6=3x^2+6x
We move all terms to the left:
x^2-x-6-(3x^2+6x)=0
We add all the numbers together, and all the variables
x^2-1x-(3x^2+6x)-6=0
We get rid of parentheses
x^2-3x^2-1x-6x-6=0
We add all the numbers together, and all the variables
-2x^2-7x-6=0
a = -2; b = -7; c = -6;
Δ = b2-4ac
Δ = -72-4·(-2)·(-6)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*-2}=\frac{6}{-4} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*-2}=\frac{8}{-4} =-2 $
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