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3x^2-10x-169=0
a = 3; b = -10; c = -169;
Δ = b2-4ac
Δ = -102-4·3·(-169)
Δ = 2128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2128}=\sqrt{16*133}=\sqrt{16}*\sqrt{133}=4\sqrt{133}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{133}}{2*3}=\frac{10-4\sqrt{133}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{133}}{2*3}=\frac{10+4\sqrt{133}}{6} $
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