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(3x+10)2x=360
We move all terms to the left:
(3x+10)2x-(360)=0
We multiply parentheses
6x^2+20x-360=0
a = 6; b = 20; c = -360;
Δ = b2-4ac
Δ = 202-4·6·(-360)
Δ = 9040
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9040}=\sqrt{16*565}=\sqrt{16}*\sqrt{565}=4\sqrt{565}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{565}}{2*6}=\frac{-20-4\sqrt{565}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{565}}{2*6}=\frac{-20+4\sqrt{565}}{12} $
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