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6(3x^2+4)-10=12(4x+16)
We move all terms to the left:
6(3x^2+4)-10-(12(4x+16))=0
We multiply parentheses
18x^2-(12(4x+16))+24-10=0
We calculate terms in parentheses: -(12(4x+16)), so:We add all the numbers together, and all the variables
12(4x+16)
We multiply parentheses
48x+192
Back to the equation:
-(48x+192)
18x^2-(48x+192)+14=0
We get rid of parentheses
18x^2-48x-192+14=0
We add all the numbers together, and all the variables
18x^2-48x-178=0
a = 18; b = -48; c = -178;
Δ = b2-4ac
Δ = -482-4·18·(-178)
Δ = 15120
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{15120}=\sqrt{144*105}=\sqrt{144}*\sqrt{105}=12\sqrt{105}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-12\sqrt{105}}{2*18}=\frac{48-12\sqrt{105}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+12\sqrt{105}}{2*18}=\frac{48+12\sqrt{105}}{36} $
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