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11x^2=4+12x
We move all terms to the left:
11x^2-(4+12x)=0
We add all the numbers together, and all the variables
11x^2-(12x+4)=0
We get rid of parentheses
11x^2-12x-4=0
a = 11; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·11·(-4)
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{5}}{2*11}=\frac{12-8\sqrt{5}}{22} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{5}}{2*11}=\frac{12+8\sqrt{5}}{22} $
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