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3x(3x-4)=51
We move all terms to the left:
3x(3x-4)-(51)=0
We multiply parentheses
9x^2-12x-51=0
a = 9; b = -12; c = -51;
Δ = b2-4ac
Δ = -122-4·9·(-51)
Δ = 1980
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1980}=\sqrt{36*55}=\sqrt{36}*\sqrt{55}=6\sqrt{55}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{55}}{2*9}=\frac{12-6\sqrt{55}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{55}}{2*9}=\frac{12+6\sqrt{55}}{18} $
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