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12x(x+3)=4(3x+1)+32
We move all terms to the left:
12x(x+3)-(4(3x+1)+32)=0
We multiply parentheses
12x^2+36x-(4(3x+1)+32)=0
We calculate terms in parentheses: -(4(3x+1)+32), so:We get rid of parentheses
4(3x+1)+32
We multiply parentheses
12x+4+32
We add all the numbers together, and all the variables
12x+36
Back to the equation:
-(12x+36)
12x^2+36x-12x-36=0
We add all the numbers together, and all the variables
12x^2+24x-36=0
a = 12; b = 24; c = -36;
Δ = b2-4ac
Δ = 242-4·12·(-36)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-48}{2*12}=\frac{-72}{24} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+48}{2*12}=\frac{24}{24} =1 $
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