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(2x+2)(3x+1)+52=180
We move all terms to the left:
(2x+2)(3x+1)+52-(180)=0
We add all the numbers together, and all the variables
(2x+2)(3x+1)-128=0
We multiply parentheses ..
(+6x^2+2x+6x+2)-128=0
We get rid of parentheses
6x^2+2x+6x+2-128=0
We add all the numbers together, and all the variables
6x^2+8x-126=0
a = 6; b = 8; c = -126;
Δ = b2-4ac
Δ = 82-4·6·(-126)
Δ = 3088
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3088}=\sqrt{16*193}=\sqrt{16}*\sqrt{193}=4\sqrt{193}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{193}}{2*6}=\frac{-8-4\sqrt{193}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{193}}{2*6}=\frac{-8+4\sqrt{193}}{12} $
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