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3x+(x+12)+2x(2x+4)=360
We move all terms to the left:
3x+(x+12)+2x(2x+4)-(360)=0
We multiply parentheses
4x^2+3x+(x+12)+8x-360=0
We get rid of parentheses
4x^2+3x+x+8x+12-360=0
We add all the numbers together, and all the variables
4x^2+12x-348=0
a = 4; b = 12; c = -348;
Δ = b2-4ac
Δ = 122-4·4·(-348)
Δ = 5712
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5712}=\sqrt{16*357}=\sqrt{16}*\sqrt{357}=4\sqrt{357}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{357}}{2*4}=\frac{-12-4\sqrt{357}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{357}}{2*4}=\frac{-12+4\sqrt{357}}{8} $
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