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2x^2-29x-49=0
a = 2; b = -29; c = -49;
Δ = b2-4ac
Δ = -292-4·2·(-49)
Δ = 1233
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1233}=\sqrt{9*137}=\sqrt{9}*\sqrt{137}=3\sqrt{137}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-3\sqrt{137}}{2*2}=\frac{29-3\sqrt{137}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+3\sqrt{137}}{2*2}=\frac{29+3\sqrt{137}}{4} $
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