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6x^2+25x+14=13
We move all terms to the left:
6x^2+25x+14-(13)=0
We add all the numbers together, and all the variables
6x^2+25x+1=0
a = 6; b = 25; c = +1;
Δ = b2-4ac
Δ = 252-4·6·1
Δ = 601
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{-(25)-\sqrt{601}}{2*6}=\frac{-25-\sqrt{601}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{601}}{2*6}=\frac{-25+\sqrt{601}}{12} $
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