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12x^2=8x+28
We move all terms to the left:
12x^2-(8x+28)=0
We get rid of parentheses
12x^2-8x-28=0
a = 12; b = -8; c = -28;
Δ = b2-4ac
Δ = -82-4·12·(-28)
Δ = 1408
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{22}}{2*12}=\frac{8-8\sqrt{22}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{22}}{2*12}=\frac{8+8\sqrt{22}}{24} $
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