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3x^2-6=17x
We move all terms to the left:
3x^2-6-(17x)=0
a = 3; b = -17; c = -6;
Δ = b2-4ac
Δ = -172-4·3·(-6)
Δ = 361
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-19}{2*3}=\frac{-2}{6} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+19}{2*3}=\frac{36}{6} =6 $
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