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4x+20=x^2-12
We move all terms to the left:
4x+20-(x^2-12)=0
We get rid of parentheses
-x^2+4x+12+20=0
We add all the numbers together, and all the variables
-1x^2+4x+32=0
a = -1; b = 4; c = +32;
Δ = b2-4ac
Δ = 42-4·(-1)·32
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*-1}=\frac{-16}{-2} =+8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*-1}=\frac{8}{-2} =-4 $
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