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3x^2-8x=48
We move all terms to the left:
3x^2-8x-(48)=0
a = 3; b = -8; c = -48;
Δ = b2-4ac
Δ = -82-4·3·(-48)
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{10}}{2*3}=\frac{8-8\sqrt{10}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{10}}{2*3}=\frac{8+8\sqrt{10}}{6} $
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