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(7x+3)(4x+1)+65=180
We move all terms to the left:
(7x+3)(4x+1)+65-(180)=0
We add all the numbers together, and all the variables
(7x+3)(4x+1)-115=0
We multiply parentheses ..
(+28x^2+7x+12x+3)-115=0
We get rid of parentheses
28x^2+7x+12x+3-115=0
We add all the numbers together, and all the variables
28x^2+19x-112=0
a = 28; b = 19; c = -112;
Δ = b2-4ac
Δ = 192-4·28·(-112)
Δ = 12905
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{12905}}{2*28}=\frac{-19-\sqrt{12905}}{56} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{12905}}{2*28}=\frac{-19+\sqrt{12905}}{56} $
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