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x^2+2x-16=12x
We move all terms to the left:
x^2+2x-16-(12x)=0
We add all the numbers together, and all the variables
x^2-10x-16=0
a = 1; b = -10; c = -16;
Δ = b2-4ac
Δ = -102-4·1·(-16)
Δ = 164
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{41}}{2*1}=\frac{10-2\sqrt{41}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{41}}{2*1}=\frac{10+2\sqrt{41}}{2} $
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