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x^2+2=3(x+4)
We move all terms to the left:
x^2+2-(3(x+4))=0
We calculate terms in parentheses: -(3(x+4)), so:We get rid of parentheses
3(x+4)
We multiply parentheses
3x+12
Back to the equation:
-(3x+12)
x^2-3x-12+2=0
We add all the numbers together, and all the variables
x^2-3x-10=0
a = 1; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·1·(-10)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*1}=\frac{-4}{2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*1}=\frac{10}{2} =5 $
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