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(3x+16)(2x+4)=180
We move all terms to the left:
(3x+16)(2x+4)-(180)=0
We multiply parentheses ..
(+6x^2+12x+32x+64)-180=0
We get rid of parentheses
6x^2+12x+32x+64-180=0
We add all the numbers together, and all the variables
6x^2+44x-116=0
a = 6; b = 44; c = -116;
Δ = b2-4ac
Δ = 442-4·6·(-116)
Δ = 4720
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4720}=\sqrt{16*295}=\sqrt{16}*\sqrt{295}=4\sqrt{295}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{295}}{2*6}=\frac{-44-4\sqrt{295}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{295}}{2*6}=\frac{-44+4\sqrt{295}}{12} $
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