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(3x-6)(x-6)=40x+240
We move all terms to the left:
(3x-6)(x-6)-(40x+240)=0
We get rid of parentheses
(3x-6)(x-6)-40x-240=0
We multiply parentheses ..
(+3x^2-18x-6x+36)-40x-240=0
We get rid of parentheses
3x^2-18x-6x-40x+36-240=0
We add all the numbers together, and all the variables
3x^2-64x-204=0
a = 3; b = -64; c = -204;
Δ = b2-4ac
Δ = -642-4·3·(-204)
Δ = 6544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6544}=\sqrt{16*409}=\sqrt{16}*\sqrt{409}=4\sqrt{409}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-4\sqrt{409}}{2*3}=\frac{64-4\sqrt{409}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+4\sqrt{409}}{2*3}=\frac{64+4\sqrt{409}}{6} $
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