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-3x^2+42x+18=0
a = -3; b = 42; c = +18;
Δ = b2-4ac
Δ = 422-4·(-3)·18
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{-(42)-6\sqrt{55}}{2*-3}=\frac{-42-6\sqrt{55}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{55}}{2*-3}=\frac{-42+6\sqrt{55}}{-6} $
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