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(2x+10)(2x+6)=132
We move all terms to the left:
(2x+10)(2x+6)-(132)=0
We multiply parentheses ..
(+4x^2+12x+20x+60)-132=0
We get rid of parentheses
4x^2+12x+20x+60-132=0
We add all the numbers together, and all the variables
4x^2+32x-72=0
a = 4; b = 32; c = -72;
Δ = b2-4ac
Δ = 322-4·4·(-72)
Δ = 2176
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{34}}{2*4}=\frac{-32-8\sqrt{34}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{34}}{2*4}=\frac{-32+8\sqrt{34}}{8} $
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