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15(6x+1)6x=183
We move all terms to the left:
15(6x+1)6x-(183)=0
We multiply parentheses
540x^2+90x-183=0
a = 540; b = 90; c = -183;
Δ = b2-4ac
Δ = 902-4·540·(-183)
Δ = 403380
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{403380}=\sqrt{324*1245}=\sqrt{324}*\sqrt{1245}=18\sqrt{1245}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-18\sqrt{1245}}{2*540}=\frac{-90-18\sqrt{1245}}{1080} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+18\sqrt{1245}}{2*540}=\frac{-90+18\sqrt{1245}}{1080} $
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