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4x(3+x)=36
We move all terms to the left:
4x(3+x)-(36)=0
We add all the numbers together, and all the variables
4x(x+3)-36=0
We multiply parentheses
4x^2+12x-36=0
a = 4; b = 12; c = -36;
Δ = b2-4ac
Δ = 122-4·4·(-36)
Δ = 720
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{5}}{2*4}=\frac{-12-12\sqrt{5}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{5}}{2*4}=\frac{-12+12\sqrt{5}}{8} $
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