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3x(10x-8)+2x(5x-12)+120=540
We move all terms to the left:
3x(10x-8)+2x(5x-12)+120-(540)=0
We add all the numbers together, and all the variables
3x(10x-8)+2x(5x-12)-420=0
We multiply parentheses
30x^2+10x^2-24x-24x-420=0
We add all the numbers together, and all the variables
40x^2-48x-420=0
a = 40; b = -48; c = -420;
Δ = b2-4ac
Δ = -482-4·40·(-420)
Δ = 69504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{69504}=\sqrt{64*1086}=\sqrt{64}*\sqrt{1086}=8\sqrt{1086}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{1086}}{2*40}=\frac{48-8\sqrt{1086}}{80} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{1086}}{2*40}=\frac{48+8\sqrt{1086}}{80} $
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