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12x+36=8x+x^2
We move all terms to the left:
12x+36-(8x+x^2)=0
We get rid of parentheses
-x^2-8x+12x+36=0
We add all the numbers together, and all the variables
-1x^2+4x+36=0
a = -1; b = 4; c = +36;
Δ = b2-4ac
Δ = 42-4·(-1)·36
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*-1}=\frac{-4-4\sqrt{10}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*-1}=\frac{-4+4\sqrt{10}}{-2} $
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