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3x^2-6x-49=0
a = 3; b = -6; c = -49;
Δ = b2-4ac
Δ = -62-4·3·(-49)
Δ = 624
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{39}}{2*3}=\frac{6-4\sqrt{39}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{39}}{2*3}=\frac{6+4\sqrt{39}}{6} $
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