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(3x+1)(4x+1)=40
We move all terms to the left:
(3x+1)(4x+1)-(40)=0
We multiply parentheses ..
(+12x^2+3x+4x+1)-40=0
We get rid of parentheses
12x^2+3x+4x+1-40=0
We add all the numbers together, and all the variables
12x^2+7x-39=0
a = 12; b = 7; c = -39;
Δ = b2-4ac
Δ = 72-4·12·(-39)
Δ = 1921
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{-(7)-\sqrt{1921}}{2*12}=\frac{-7-\sqrt{1921}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{1921}}{2*12}=\frac{-7+\sqrt{1921}}{24} $
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