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(2x+13)(3x)=47
We move all terms to the left:
(2x+13)(3x)-(47)=0
We multiply parentheses
6x^2+39x-47=0
a = 6; b = 39; c = -47;
Δ = b2-4ac
Δ = 392-4·6·(-47)
Δ = 2649
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{-(39)-\sqrt{2649}}{2*6}=\frac{-39-\sqrt{2649}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{2649}}{2*6}=\frac{-39+\sqrt{2649}}{12} $
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