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40x-3x^2=44
We move all terms to the left:
40x-3x^2-(44)=0
a = -3; b = 40; c = -44;
Δ = b2-4ac
Δ = 402-4·(-3)·(-44)
Δ = 1072
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1072}=\sqrt{16*67}=\sqrt{16}*\sqrt{67}=4\sqrt{67}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{67}}{2*-3}=\frac{-40-4\sqrt{67}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{67}}{2*-3}=\frac{-40+4\sqrt{67}}{-6} $
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