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x^2+12x+20=3x
We move all terms to the left:
x^2+12x+20-(3x)=0
We add all the numbers together, and all the variables
x^2+9x+20=0
a = 1; b = 9; c = +20;
Δ = b2-4ac
Δ = 92-4·1·20
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*1}=\frac{-10}{2} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*1}=\frac{-8}{2} =-4 $
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