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2x^2-42x+147=0
a = 2; b = -42; c = +147;
Δ = b2-4ac
Δ = -422-4·2·147
Δ = 588
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14\sqrt{3}}{2*2}=\frac{42-14\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{3}}{2*2}=\frac{42+14\sqrt{3}}{4} $
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