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3x^2-48x+100=0
a = 3; b = -48; c = +100;
Δ = b2-4ac
Δ = -482-4·3·100
Δ = 1104
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1104}=\sqrt{16*69}=\sqrt{16}*\sqrt{69}=4\sqrt{69}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-4\sqrt{69}}{2*3}=\frac{48-4\sqrt{69}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+4\sqrt{69}}{2*3}=\frac{48+4\sqrt{69}}{6} $
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