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(12x-3)(5x+12)=(3x+13)
We move all terms to the left:
(12x-3)(5x+12)-((3x+13))=0
We multiply parentheses ..
(+60x^2+144x-15x-36)-((3x+13))=0
We calculate terms in parentheses: -((3x+13)), so:We get rid of parentheses
(3x+13)
We get rid of parentheses
3x+13
Back to the equation:
-(3x+13)
60x^2+144x-15x-3x-36-13=0
We add all the numbers together, and all the variables
60x^2+126x-49=0
a = 60; b = 126; c = -49;
Δ = b2-4ac
Δ = 1262-4·60·(-49)
Δ = 27636
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{27636}=\sqrt{196*141}=\sqrt{196}*\sqrt{141}=14\sqrt{141}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(126)-14\sqrt{141}}{2*60}=\frac{-126-14\sqrt{141}}{120} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(126)+14\sqrt{141}}{2*60}=\frac{-126+14\sqrt{141}}{120} $
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