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2x+(2x+2)(2x-4)=60
We move all terms to the left:
2x+(2x+2)(2x-4)-(60)=0
We multiply parentheses ..
(+4x^2-8x+4x-8)+2x-60=0
We get rid of parentheses
4x^2-8x+4x+2x-8-60=0
We add all the numbers together, and all the variables
4x^2-2x-68=0
a = 4; b = -2; c = -68;
Δ = b2-4ac
Δ = -22-4·4·(-68)
Δ = 1092
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1092}=\sqrt{4*273}=\sqrt{4}*\sqrt{273}=2\sqrt{273}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{273}}{2*4}=\frac{2-2\sqrt{273}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{273}}{2*4}=\frac{2+2\sqrt{273}}{8} $
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