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4x^2+4x-16=6x-12
We move all terms to the left:
4x^2+4x-16-(6x-12)=0
We get rid of parentheses
4x^2+4x-6x+12-16=0
We add all the numbers together, and all the variables
4x^2-2x-4=0
a = 4; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·4·(-4)
Δ = 68
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{17}}{2*4}=\frac{2-2\sqrt{17}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{17}}{2*4}=\frac{2+2\sqrt{17}}{8} $
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