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(2x+10)(2x+6)=120
We move all terms to the left:
(2x+10)(2x+6)-(120)=0
We multiply parentheses ..
(+4x^2+12x+20x+60)-120=0
We get rid of parentheses
4x^2+12x+20x+60-120=0
We add all the numbers together, and all the variables
4x^2+32x-60=0
a = 4; b = 32; c = -60;
Δ = b2-4ac
Δ = 322-4·4·(-60)
Δ = 1984
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{31}}{2*4}=\frac{-32-8\sqrt{31}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{31}}{2*4}=\frac{-32+8\sqrt{31}}{8} $
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