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3x^2-18x+5=47.
We move all terms to the left:
3x^2-18x+5-(47.)=0
We add all the numbers together, and all the variables
3x^2-18x+5-47=0
We add all the numbers together, and all the variables
3x^2-18x-42=0
a = 3; b = -18; c = -42;
Δ = b2-4ac
Δ = -182-4·3·(-42)
Δ = 828
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{828}=\sqrt{36*23}=\sqrt{36}*\sqrt{23}=6\sqrt{23}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{23}}{2*3}=\frac{18-6\sqrt{23}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{23}}{2*3}=\frac{18+6\sqrt{23}}{6} $
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