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2x^2-21x=49=0
We move all terms to the left:
2x^2-21x-(49)=0
a = 2; b = -21; c = -49;
Δ = b2-4ac
Δ = -212-4·2·(-49)
Δ = 833
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{833}=\sqrt{49*17}=\sqrt{49}*\sqrt{17}=7\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-7\sqrt{17}}{2*2}=\frac{21-7\sqrt{17}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+7\sqrt{17}}{2*2}=\frac{21+7\sqrt{17}}{4} $
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