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180=30-12x+(2x^2)
We move all terms to the left:
180-(30-12x+(2x^2))=0
We get rid of parentheses
-2x^2+12x-30+180=0
We add all the numbers together, and all the variables
-2x^2+12x+150=0
a = -2; b = 12; c = +150;
Δ = b2-4ac
Δ = 122-4·(-2)·150
Δ = 1344
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{21}}{2*-2}=\frac{-12-8\sqrt{21}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{21}}{2*-2}=\frac{-12+8\sqrt{21}}{-4} $
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