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12x=5-4x+2x^2
We move all terms to the left:
12x-(5-4x+2x^2)=0
We get rid of parentheses
-2x^2+4x+12x-5=0
We add all the numbers together, and all the variables
-2x^2+16x-5=0
a = -2; b = 16; c = -5;
Δ = b2-4ac
Δ = 162-4·(-2)·(-5)
Δ = 216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{6}}{2*-2}=\frac{-16-6\sqrt{6}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{6}}{2*-2}=\frac{-16+6\sqrt{6}}{-4} $
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